CHEM 131 - ADVANCED GENERAL CHEMISTRY I Gas Stoichiometry Question Bank Set 1

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Gas Stoichiometry

Question Bank - Set 1

Liberty University

Question 1

Question

A 1.50 L container holds a mixture of nitrogen gas and oxygen gas at 27

C and

0.750 atm. If the mole fraction of nitrogen gas in the mixture is 0.600, what is

the partial pressure of oxygen gas in the container?

Solution

Step 1: Calculate the total moles of gas in the container. Given: Volume (V)

= 1.50 L, Temperature (T) = 27

C = 300 K, Pressure (P) = 0.750 atm, Mole

fraction of nitrogen gas (XN2) = 0.600.

The ideal gas equation is given by:

P V =nRT

Where nis the total moles of gas and Ris the ideal gas constant.

Plugging in the values:

n=P V

n=(0.750 atm)(1.50 L)

(0.0821 L atm/mol K)(300 K)

Calculating n, we find:

n≈0.0452 moles

Step 2: Determine the moles of nitrogen gas and oxygen gas in the mixture.

Since the mole fraction of nitrogen gas is 0.600, the mole fraction of oxygen gas

(XO2) would be:

XO2= 1 −XN2

XO2= 1 −0.600

XO2= 0.400

Therefore, the moles of nitrogen gas (nN2) and oxygen gas (nO2) can be

calculated as:

nN2=XN2×n

nN2= 0.600 ×0.0452

nN2= 0.0271 moles

nO2=XO2×n

nO2= 0.400 ×0.0452

nO2= 0.0181 moles

Step 3: Calculate the partial pressure of oxygen gas. The partial pressure

of oxygen gas (PO2) can be calculated using the formula:

PO2=nO2RT

PO2=(0.0181 moles)(0.0821 L atm/mol K)(300 K)

1.50 L

Calculating PO2, we find:

PO2≈0.0915 atm

Therefore, the partial pressure of oxygen gas in the container is approxi-

mately 0.0915 atm.

Question 2

Question

A gaseous compound containing only carbon and hydrogen is burned in air. A

1.50 g sample of the compound yields 4.20 g of carbon dioxide and 1.71 g of

water. Determine the empirical formula of the compound.

Solution

Step 1: Calculate the moles of carbon dioxide produced. Given: Mass of carbon

dioxide produced = 4.20 g

Molar mass of carbon dioxide (CO2) = 44.01 g/mol

Number of moles of CO2=4.20 g

44.01 g/mol = 0.0954 mol

Step 2: Calculate the moles of water produced. Given: Mass of water pro-

duced = 1.71 g

Molar mass of water (H2O) = 18.02 g/mol

Number of moles of H2O = 1.71 g

18.02 g/mol = 0.0949 mol

Step 3: Determine the moles of carbon and hydrogen in the compound.

From the balanced chemical equation for the combustion of the compound: 1

mol of the compound yields 1 mol of CO2and 1 mol of H2O

Therefore, the moles of carbon in the compound = 0.0954 mol The moles of

hydrogen in the compound = 0.0949 mol

Step 4: Calculate the mole ratio of carbon to hydrogen. Assume the com-

pound has the general formula CaHb, where ’a’ and ’b’ are the subscripts we

want to determine.

From the moles of carbon and hydrogen: a=0.0954

0.0949 ≈1

b=0.0949

0.0949 = 1

Thus, the empirical formula of the compound is CH.

Step 5: Write the empirical formula. The empirical formula of the compound

is CH.

Question 3

Question

A 1.00 L container holds a mixture of oxygen gas and nitrogen gas at a total

pressure of 3.00 atm and a temperature of 25

C. The partial pressure of oxygen

gas is 1.20 atm. If all the oxygen gas is consumed in a reaction that produces

nitrogen dioxide gas, what will be the partial pressure of nitrogen dioxide gas in

the container at the end of the reaction? Assume the reaction goes to completion

and the volume and temperature remain constant.

Solution

Step 1: Calculate the initial moles of oxygen gas using the ideal gas law:

n=P V

Given:

Poxygen = 1.20 atm

Ptotal = 3.00 atm

V= 1.00 L

T= 25C= 298 K

R= 0.0821 L

atm/mol

Plugging in the values:

noxygen =(1.20 atm)(1.00 L)

(0.0821 L

atm/mol

K)(298 K)

noxygen ≈0.0589 mol

Step 2: Since the reaction goes to completion without change in volume or

temperature, all the oxygen gas reacts to form nitrogen dioxide gas. Therefore,

the moles of nitrogen dioxide gas produced will be equal to the moles of oxygen

consumed.

Step 3: Calculate the mole fraction of nitrogen dioxide gas:

x=nnitrogen dioxide

ntotal

Since the total moles of gas is the sum of moles of oxygen and moles of

nitrogen dioxide:

ntotal =noxygen +nnitrogen dioxide

Given that the moles of nitrogen dioxide gas produced is equal to the moles

of oxygen consumed:

ntotal = 2noxygen

Substitute the values:

ntotal = 2(0.0589)

ntotal = 0.118 mol

Step 4: Calculate the total pressure of the mixture after the reaction:

Ptotal, final =Pnitrogen dioxide +Pnitrogen

Given that the moles of nitrogen gas remains unchanged:

Ptotal, final =Pnitrogen dioxide +Pnitrogen = (nnitrogen dioxide +nnitrogen)RT/V

Since the partial pressure of nitrogen gas is the same as the initial condition:

Pnitrogen = (nnitrogen)RT/V = (0.118 mol)0.0821 L

atm/mol

K(298 K)/1.00 L

Now, calculate the partial pressure of nitrogen dioxide gas:

Pnitrogen dioxide =Ptotal, final −Pnitrogen

Pnitrogen dioxide = 3.00 atm−(0.118 mol ×0.0821 L

atm/mol

K×298 K/1.00 L)

Pnitrogen dioxide = 3.00 atm −0.287 atm

Pnitrogen dioxide = 2.71 atm

Therefore, the partial pressure of nitrogen dioxide gas in the container at

the end of the reaction will be 2.71 atm.

Question 4

Question

A reaction occurs between 15.0 L of hydrogen gas at standard temperature and

pressure (STP) and excess oxygen gas. If the reaction produces water vapor,

what volume of water vapor is produced at STP?

Solution

Step 1: Write the balanced chemical equation for the reaction. The balanced

chemical equation for the reaction between hydrogen gas and oxygen gas to

produce water vapor is:

2H2(g) + O2(g)→2H2O(g)

Step 2: Determine the moles of hydrogen gas. Given that the volume of

hydrogen gas is 15.0 L at STP (standard temperature and pressure: 0

C and 1

atm), we can use the ideal gas law to find the number of moles of hydrogen gas:

P V =nRT

n=P V

RT =(1 atm)(15.0 L)

(0.0821 L ·atm/mol ·K)(273 K) = 0.681 moles

Step 3: Determine the volume of water vapor produced. From the balanced

chemical equation, we see that the ratio of moles of hydrogen gas to moles of

water vapor is 2:2. Therefore, the moles of water vapor produced will also be

0.681. Now, we can use this mole value to find the volume of water vapor

produced at STP:

P V =nRT

V=nRT

P=(0.681 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm = 15.2 L

Therefore, 15.2 L of water vapor is produced at STP.

Question 5

Question

A sample of iron metal reacts with sulfuric acid according to the following

balanced chemical equation:

Fe + H2SO4→FeSO4+ H2

If 5.00 grams of iron react with an excess of sulfuric acid, how many grams

of hydrogen gas are produced?

(Note: molar mass of iron = 55.85 g/mol, molar mass of sulfuric acid =

98.08 g/mol, molar mass of hydrogen gas = 2.02 g/mol)

Solution

Step 1: Calculate the number of moles of iron (Fe) given 5.00 grams of iron.

Moles of Fe = Mass

Molar Mass =5.00 g

55.85 g/mol ≈0.0894 mol

Step 2: Use the balanced chemical equation to determine the mole ratio be-

tween iron (Fe) and hydrogen gas (H¡sub¿2¡/sub¿). From the balanced chemical

equation, 1 mole of Fe produces 1 mole of H¡sub¿2¡/sub¿.

Step 3: Calculate the number of moles of hydrogen gas produced.

Moles of H2= Moles of Fe ×Mole Ratio = 0.0894 mol ×1 = 0.0894 mol

Step 4: Calculate the mass of hydrogen gas produced.

Mass of H2= Moles ×Molar Mass = 0.0894 mol ×2.02 g/mol = 0.181 g

Therefore, 0.181 grams of hydrogen gas are produced when 5.00 grams of

iron react with sulfuric acid.

Question 6

Question

A gaseous compound contains only carbon and hydrogen. When 0.500 mol of

the compound is reacted with excess oxygen, it produces 1.50 mol of carbon

dioxide and 2.00 mol of water vapor. Determine the molecular formula of the

compound.

Solution

Step 1: Write the balanced chemical equation for the reaction.

CxHy+ O2→CO2+ H2O

Step 2: Write the stoichiometry ratio from the given information. - 0.500 mol

of the compound produces 1.50 mol of carbon dioxide. This implies a 1:3 ratio

between the compound and carbon dioxide. - 0.500 mol of the compound pro-

duces 2.00 mol of water vapor. This implies a 1:4 ratio between the compound

and water vapor.

Step 3: Determine the molar masses of carbon dioxide (44 g/mol) and

water (18 g/mol). - The compound’s molar mass can be represented as x×

molar mass of C + y×molar mass of H.

Step 4: Determine the molar mass of the compound. - From the stoichiom-

etry ratios, we can construct a system of equations to find the values of x and

Step 5: Solve for x and y. - From the system of equations, we find that x =

4 and y = 10.

Step 6: Determine the molecular formula of the compound. - The compound

is C4H10, which is the molecular formula for butane.

Question 7

Question

A mixture of nitrogen gas (N2) and oxygen gas (O2) is held in a 5.00 L container

at a temperature of 300 K. The pressure of the mixture is measured to be 2.00

atm. If the mole fraction of nitrogen gas in the mixture is 0.60, calculate the

number of moles of oxygen gas in the container.

Solution

Step 1: Calculate the total moles of gas in the container.

Given that the mole fraction of nitrogen gas is 0.60, we can calculate the moles

of nitrogen gas in the container as follows:

Moles of N2= Mole fraction ×Total moles of gas

Moles of N2= 0.60 ×Total moles of gas

Step 2: Use the ideal gas law to find the total moles of gas.

The ideal gas law is given by:

P V =nRT

where: - P is the pressure in atm, - V is the volume in liters, - n is the number of

moles, - R is the ideal gas constant (0.0821 L·atm/mol·K), - T is the temperature

in Kelvin.

We can rearrange this equation to solve for n:

n=P V

Step 3: Substitute known values into the equation to find the total moles of

gas.

n=(2.00 atm) ×(5.00 L)

(0.0821 L·atm/mol·K) ×(300 K)

n≈10.00

24.63

n≈0.41 moles

Step 4: Calculate the moles of oxygen gas in the container.

Since the moles of nitrogen gas is 0.60 times the total moles of gas:

Moles of O2= (1 −0.60) ×Total moles of gas

Moles of O2= 0.40 ×0.41

Moles of O2= 0.164 moles

Therefore, there are 0.164 moles of oxygen gas in the 5.00 L container at 300

Question 8

Question

A reaction takes place between 2.50 L of hydrogen gas at STP and an excess of

oxygen gas. If the reaction yields water vapor as the only product, how many

grams of water vapor are produced?

Solution

Step 1: Determine the moles of hydrogen gas using the ideal gas law equation:

P V =nRT

Where: - P= pressure = 1 atm - V= volume = 2.50 L - n= moles - R= ideal

gas constant = 0.0821 L

atm/K

mol - T= temperature (STP) = 273 K

Substitute the values into the equation:

1 atm ×2.50 L = n×0.0821 L

atm/K

mol ×273 K

n=1×2.50

0.0821 ×273 ≈0.105 mol

Step 2: Using the balanced chemical equation for the reaction:

2 H2+ O2→2 H2O

1 mol of H2reacts to produce 2 mol of H2O.

Therefore, the moles of water vapor produced will be:

0.105 mol ×2 mol H2O

2 mol H2

= 0.105 mol H2O

Step 3: Calculate the molar mass of water (H2O):

2×1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol

Step 4: Finally, determine the mass of water vapor produced:

0.105 mol H2O×18.02 g/mol = 1.89 g

Question 9

Question

A reaction between solid potassium chlorate and solid manganese dioxide pro-

duces potassium chloride, manganese(IV) oxide, and oxygen gas.

How many liters of oxygen gas at STP are produced when 25 g of potassium

chlorate reacts completely, assuming the reaction takes place at STP conditions?

Solution

Step 1: Write the balanced chemical equation for the reaction.

2KClO3(s)→2KCl(s) + 3O2(g)

Step 2: Calculate the molar mass of potassium chlorate (KClO).

K: 39.10 g/mol, Cl: 35.45 g/mol, O: 16.00 g/mol

Molar mass of KClO3= 39.10 + 35.45 + 3(16.00) = 122.55 g/mol

Step 3: Calculate the number of moles of potassium chlorate used.

Moles = Mass

Molar mass =25 g

122.55 g/mol = 0.204mol

Step 4: Use the balanced chemical equation to find the moles of oxygen gas

produced.

Moles of O2= 3 ×moles of KClO3= 3 ×0.204 mol = 0.612 mol

Step 5: Calculate the volume of oxygen gas at STP using the ideal gas law

(V=nRT

P).

R=0.0821 L·atm

K·mol,T = 273 K,P = 1 atm

V=0.612 mol ×0.0821 L·atm

K·mol ×273 K

1 atm = 13.7 L

So, 13.7 liters of oxygen gas at STP are produced when 25 g of potassium

chlorate reacts completely.

Question 10

Question

A reaction takes place between acetylene gas (C2H2) and oxygen gas according

to the following balanced chemical equation:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

If 10.0 L of acetylene gas at STP reacts completely with excess oxygen gas, what

volume of carbon dioxide is produced at STP?

Solution

Step 1: Find the number of moles of acetylene gas. Given that the volume of

acetylene gas is 10.0 L at STP, we can use the equation n=P V

RT to find the

number of moles.

For C2H2:nC2H2=(1.00 atm)(10.0 L)

(0.0821 L ·atm/K ·mol)(273 K) = 0.417 mol

Step 2: Determine the limiting reactant. From the balanced chemical equa-

tion, we can see that 2 moles of C2H2react with 5 moles of O2.

Moles of O2=5

2×moles of C2H2=5

2×0.417 = 1.04 mol

Step 3: Calculate the volume of carbon dioxide produced. From the balanced

chemical equation, we see that 2 moles of C2H2produce 4 moles of CO2.

Moles of CO2= 4 ×moles of C2H2= 4 ×0.417 = 1.67 mol

Now, we can calculate the volume of carbon dioxide produced using the ideal

gas law.

V=nRT

P=(1.67 mol)(0.0821 L ·atm/K ·mol)(273 K)

1.00 atm = 38.1 L

Therefore, 38.1 L of carbon dioxide is produced at STP.

Question 11

Question

A sample of sodium azide, NaN3, is heated and its decomposition reaction

produces nitrogen gas, sodium metal, and sodium oxide. If 5.00 L of nitrogen

gas is collected at 750 mmHg and 25

C, how many grams of sodium azide were

decomposed? (Assume the reaction goes to completion.)

Solution

Step 1: Write the balanced chemical equation for the decomposition of sodium

azide:

2NaN3→2Na + 3N2

Step 2: Calculate the number of moles of nitrogen gas collected using the

ideal gas law:

P V =nRT

n=P V

RT =(750 ×10−3atm) ×(5.00 L)

0.0821 L·atm

mol·K×(25 + 273) K

n= 0.0918 mol N2

Step 3: Use the balanced chemical equation to determine the moles of NaN3

decomposed: Since the molar ratio between NaN3and N2is 2:3, the moles of

NaN3decomposed is calculated as follows:

moles of NaN3=2

3×0.0918mol = 0.0612mol NaN3

Step 4: Calculate the molar mass of NaN3:

Molar mass of NaN3= 3(22.99) + 14.01 + 3(16.00) = 65.02 g/mol

Step 5: Determine the mass of sodium azide decomposed:

Mass of NaN3= 0.0612mol ×65.02 g/mol = 3.98 g

Therefore, 3.98 grams of sodium azide were decomposed.

Question 12

Question

A 2.50 L container at 27

C contains a mixture of 2.63 g of helium (He) and 8.65

g of neon (Ne) at a total pressure of 1.25 atm. Assuming ideal gas behavior,

determine the partial pressure of each gas in the mixture.

Solution

Step 1: Calculate the moles of helium and neon present. Step 2: Determine the

mole fraction of each gas. Step 3: Calculate the partial pressure of each gas

using the mole fractions.

Step 1: Calculate the moles of helium and neon present. The molar mass

of helium (He) is 4.00 g/mol, and the molar mass of neon (Ne) is 20.18 g/mol.

Number of moles of helium:

nHe =2.63 g

4.00 g/mol = 0.6575 mol

Number of moles of neon:

nNe =8.65 g

20.18 g/mol = 0.4281 mol

Step 2: Determine the mole fraction of each gas. Total moles of gas:

ntotal =nHe +nNe = 0.6575 mol + 0.4281 mol = 1.0856 mol

Mole fraction of helium:

XHe =nHe

ntotal

=0.6575 mol

1.0856 mol = 0.6060

Mole fraction of neon:

XNe =nNe

ntotal

=0.4281 mol

1.0856 mol = 0.3940

Step 3: Calculate the partial pressure of each gas using the mole fractions.

Partial pressure of helium:

PHe =XHe ·Ptotal = 0.6060 ×1.25 atm = 0.7575 atm

Partial pressure of neon:

PNe =XNe ·Ptotal = 0.3940 ×1.25 atm = 0.4925 atm

Therefore, the partial pressure of helium is 0.7575 atm and the partial pres-

sure of neon is 0.4925 atm in the mixture.

Question 13

Question

A gaseous compound contains only carbon, hydrogen, and oxygen. When a

0.500 g sample of the compound is completely combusted in excess oxygen,

1.100 g of carbon dioxide and 0.450 g of water are produced. Determine the

empirical formula of the compound.

(Given: molar mass of carbon dioxide = 44.01 g/mol, molar mass of water

= 18.015 g/mol)

Solution

Step 1: Calculate the moles of carbon dioxide produced. Let x be the moles of

carbon in the compound.

Moles of CO2=1.100 g

44.01 g/mol = 0.025 mol

Step 2: Calculate the moles of water produced. From the given information,

we can calculate the moles of hydrogen in the compound. Moles of hydrogen in

water = 0.450 g

18.015 g/mol = 0.025 mol

Step 3: Calculate the moles of oxygen in the compound. Let y be the moles

of oxygen in the compound. From the combustion reaction, the moles of oxygen

can be determined based on the moles of carbon and hydrogen. The moles of

oxygen can be calculated using the mole ratios in the compound.

Step 4: Determine the empirical formula of the compound. The ratio of

moles of carbon, hydrogen, and oxygen in the compound can be used to write

the empirical formula.

Thus, the empirical formula of the compound is determined to be C1H1O1,

which simplifies to CH2O.

Question 14

Question

A mixture of methane gas (CH4) and oxygen gas (O2) is combusted in a closed

container. If 15.0 g of methane reacts with excess oxygen to produce carbon

dioxide and water, determine the volume of carbon dioxide produced at STP.

Assume complete combustion.

Solution

Step 1: Write the balanced chemical equation for the combustion of methane.

CH4+ 2O2→CO2+ 2H2O

Step 2: Calculate the number of moles of methane used in the reaction.

Given: Mass of methane = 15.0 g Molar mass of methane (CH4) = 16.04

g/mol

Number of moles of methane = 15.0 g

16.04 g/mol = 0.9351 mol

Step 3: Determine the number of moles of carbon dioxide produced using

the stoichiometry of the balanced chemical equation.

From the balanced equation, 1 mol of methane produces 1 mol of carbon

dioxide. Number of moles of carbon dioxide = 0.9351 mol

Step 4: Calculate the volume of carbon dioxide produced at STP.

Given: Standard Temperature and Pressure (STP) conditions are 0

C (273 K)

and 1 atm pressure. Molar volume of gas at STP = 22.4 L/mol

Volume of carbon dioxide = 0.9351 mol ×22.4 L/mol = 20.95 L

Answer: The volume of carbon dioxide produced at STP is 20.95 L.

Question 15

Question

Methane (CH4) undergoes combustion according to the following balanced

chemical equation:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

If 8.00 L of methane is reacted with excess oxygen, how many moles of water

vapor will be produced? Assume all gases are at the same temperature and

pressure.

Solution

Step 1: Write down the given information and the balanced chemical equation.

Given: Volume of CH4= 8.00LCoefficients of H2O(g) in the balanced chemical

equation = 2

Balanced chemical equation:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Step 2: Find the number of moles of CH4. Given volume of CH4= 8.00L

Using the ideal gas law, we can convert the volume to moles:

n=P V

Step 3: Calculate the number of moles of H2Oproduced. From the balanced

chemical equation, we see that 1 mole of CH4produces 2 moles of H2O.

So, moles of H2Oproduced = moles of CH4×2

Step 4: Substitute the values into the equations and solve for the number of

moles of H2Oproduced.

n=P V

nCH4=(1)(8.00)

(0.0821)(298)

nCH4≈0.326 moles

Moles of H2Oproduced = 0.326 ×2=0.652 moles

Therefore, 0.652 moles of water vapor will be produced when 8.00 L of

methane undergoes combustion.

Question 16

Question

A reaction between solid magnesium and hydrochloric acid produces hydrogen

gas. If 3.00 g of magnesium reacts completely with an excess of hydrochloric

acid to produce 2.15 L of hydrogen gas at a temperature of 25

C and a pressure

of 1.00 atm, determine the balanced chemical equation for the reaction and

the volume of hydrogen gas at STP that would be produced from 4.50 g of

magnesium.

Solution

Step 1: Determine the balanced chemical equation for the reaction between

magnesium and hydrochloric acid.

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the number of moles of hydrogen gas produced in the

reaction with 3.00 g of magnesium.

Moles of Mg = 3.00 g

24.305 g/mol = 0.1235 mol

Moles of H2= 0.1235 mol ×1 mol H2

1 mol Mg = 0.1235 mol

Step 3: Use the ideal gas law to determine the volume of hydrogen gas

produced at STP.

PV = nRT

(1.00 atm)(2.15 L) = (0.1235 mol)(0.0821 L ·atm/mol ·K)(298 K)

Volume of H2=0.1235 ×0.0821 ×298 ×2.15

1.00 ≈6.69 L at STP

Step 4: Determine the volume of hydrogen gas produced at STP from 4.50

g of magnesium.

Moles of Mg = 4.50 g

24.305 g/mol = 0.1853 mol

Moles of H2= 0.1853 mol ×1 mol H2

1 mol Mg = 0.1853 mol

Volume of H2=0.1853 ×0.0821 ×298 ×2.15

1.00 ≈8.23 L at STP

Therefore, the volume of hydrogen gas produced at STP from 4.50 g of

magnesium is approximately 8.23 L.

Question 17

Question

A 2.00 L container at 25

C is filled with 0.400 mol of nitrogen gas and 1.00

mol of oxygen gas. If the two gases react according to the following balanced

equation:

N2(g)+O2(g)→2NO(g)

What is the total pressure in the container at 25

C after the reaction is complete?

Assume ideal gas behavior.

Solution

Step 1: Write down the balanced chemical equation.

N2(g)+O2(g)→2NO(g)

Step 2: Calculate the moles of nitrogen and oxygen gas initially. - Moles of

nitrogen gas, nN2= 0.400 mol - Moles of oxygen gas, nO2= 1.00 mol

Step 3: Determine the limiting reactant by comparing the mole ratios in the

balanced equation. Since the stoichiometry of the balanced equation is 1:1 for

N2and O2, oxygen gas is the limiting reactant.

Step 4: Use the ideal gas law to find the partial pressure of each gas before

the reaction. - For nitrogen gas: PN2=nN2RT

V=(0.400mol)(0.0821 atm·L

mol·K)(298K)

2.00L=

4.94atm - For oxygen gas: PO2=nO2RT

V=(1.00mol)(0.0821 atm·L

mol·K)(298K)

2.00L= 12.4atm

Step 5: Determine the change in moles for the gases. - For nitrogen gas:

∆nN2=−1(0.400) = −0.400 mol - For oxygen gas: ∆nO2=−1(1.00) = −1.00

mol - For nitrogen monoxide gas: ∆nN O = +2(1.00) = 2.00 mol

Step 6: Calculate the total moles of gas at the end. - Total moles of gas

=nN2+nO2+ ∆nNO = 0.400 + 1.00 + 2.00 = 3.40 mol

Step 7: Use the ideal gas law again to find the total pressure in the container

at the end.

Ptotal =ntotalRT

V=(3.40mol)(0.0821 atm·L

mol·K)(298K)

2.00L= 14.2atm

Therefore, the total pressure in the container at 25

C after the reaction is

complete is 14.2 atm.

Question 18

Question

A 2.50 L container at 25

C and 1.00 atm contains 0.500 mol of nitrogen gas,

1.00 mol of oxygen gas, and 2.00 mol of argon gas. If the container is heated to

160

C, what will be the new pressure inside the container?

Solution

Step 1: Calculate the initial total pressure inside the container using the ideal

gas law:

P V =nRT

Given: Initial volume (V) = 2.50 L, Initial temperature (T) = 25C= 25 +

273.15 K, Initial pressure (P)=1.00 atm, Moles of nitrogen (nN2)=0.500 mol,

Moles of oxygen (nO2) = 1.00 mol, Moles of argon (nAr)=2.00 mol.

First, let’s find the total moles of gas in the container:

ntotal =nN2+nO2+nAr

ntotal = 0.500 + 1.00 + 2.00 = 3.50 mol

Now, substitute the given values into the ideal gas law to find the initial

pressure:

Pinitial =ntotalRT

Pinitial =3.50 ×0.0821 ×(25 + 273.15)

2.50

Pinitial =3.50 ×0.0821 ×298.15

2.50

Pinitial =86.487

2.50

Pinitial = 34.59 atm

Step 2: Calculate the final pressure inside the container after heating the

gases to 160

C using the ideal gas law:

Pfinal =ntotalRT

Substitute the new temperature (T= 160 + 273.15 K) into the ideal gas law

to find the final pressure:

Pfinal =3.50 ×0.0821 ×(160 + 273.15)

2.50

Pfinal =3.50 ×0.0821 ×433.15

2.50

Pfinal =123.929

2.50

Pfinal = 49.57 atm

Therefore, the new pressure inside the container after heating the gases to

160

C will be 49.57 atm.

Question 19

Question

A 2.00 L container holds a mixture of H2, O2, and H2O gases at 425◦C and

1.00 atm. If the total pressure in the container is 3.00 atm, what is the partial

pressure of each gas?

Solution

Step 1: Calculate the moles of each gas based on the ideal gas law, P V =nRT ,

where Pis the pressure, Vis the volume, nis the number of moles, Ris the

gas constant, and Tis the temperature in Kelvin.

Given: P= 1.00 atm = 1.013 ×105Pa V= 2.00 L T= 425◦C = (425 +273)

K = 698 K R= 8.314 J/(mol·K)

For H2: 1.00 ×2.00 = n×8.314 ×698 n=1.00×2.00

8.314×698

Step 2: Repeat the calculation for O2and H2O.

nO2 =2.00×2.00

8.314×698 nH2O =(3.00−1.00)×2.00

8.314×698

Step 3: Calculate the mole fractions of each gas.

Xi=ni

ntotal

Step 4: Calculate the partial pressure of each gas using Dalton’s law of

partial pressures, Pi=Xi·Ptotal.

PH2 =XH2 ×1.00 PO2 =XO2 ×1.00 PH2O =XH2O ×2.00

Question 20

Question

A gas mixture contains helium and argon. If 8.00 L of the mixture at 27

C and

1.00 atm pressure contains 0.500 mol of helium and 1.00 mol of argon, what is

the partial pressure of each gas in the mixture?

Solution

Step 1: Calculate the total number of moles in the mixture. Let ntotal be the

total number of moles in the mixture. We are given that there are 0.500 mol of

helium and 1.00 mol of argon.

ntotal = 0.500 mol + 1.00 mol = 1.50 mol

Step 2: Calculate the mole fraction of each gas. The mole fraction of helium,

denoted by XHe, is defined as the moles of helium divided by the total moles in

the mixture. Similarly, the mole fraction of argon, denoted by XAr, is given by

the moles of argon divided by the total moles in the mixture.

XHe =0.500 mol

1.50 mol =1

XAr =1.00 mol

1.50 mol =2

Step 3: Calculate the partial pressure of each gas. The partial pressure of a

gas in a mixture is given by the product of the mole fraction of the gas and the

total pressure of the mixture.

PHe =XHe ×Ptotal =1

3×1.00 atm = 1

3atm

PAr =XAr ×Ptotal =2

3×1.00 atm = 2

3atm

Therefore, the partial pressure of helium is 1

3atm and the partial pressure

of argon is 2

3atm in the mixture.

Question 21

Question

A chemical reaction produces 8.60 L of carbon dioxide gas (CO2) at a temper-

ature of 25

C and pressure of 2.00 atm. If the reaction is known to produce

1.5 moles of CO2, what is the volume of oxygen gas (O2) produced at the same

conditions?

(Note: Assume all gases are ideal and use standard temperature and pressure

(STP).)

Solution

Step 1: Calculate the number of moles of CO2produced using the ideal gas law

formula:

PV = nRT

Given that the pressure (P) = 2.00 atm, volume (V) = 8.60 L, temperature

(T) = 25

C (which is 298 K), and we are solving for the number of moles (n):

nCO2=PV

RT =(2.00 atm)(8.60 L)

0.0821 L atm/mol K ×298 K

nCO2= 0.718 mol

Step 2: Use the stoichiometry of the reaction to find the number of moles

of O2produced. Since we know that 1.5 moles of CO2produces x moles of O2,

we can set up the following mole ratio:

nO2

nCO2

=1.5 mol CO2

1 mol O2

Substitute in the value of nCO2from step 1:

nO2

0.718 mol CO2

=1.5 mol CO2

1 mol O2

nO2= 1.078 mol O2

Step 3: Calculate the volume of O2gas produced using the ideal gas law

formula with the same temperature and pressure:

PV = nRT

VO2=nO2×RT

P=(1.078 mol)(0.0821 L atm/mol K ×298 K)

2.00 atm

VO2= 13.2 L

Therefore, the volume of O2gas produced at the same conditions is 13.2 L.

Question 22

Question

A 2.00 L container holds 1.00 g of nitrogen gas and 1.50 g of oxygen gas at 25

What is the partial pressure of each gas, and what is the total pressure in the

container?

Given: Molar mass of N2= 28.0 g/mol Molar mass of O2= 32.0 g/mol

Universal gas constant, R = 0.0821 L atm mol−1K−1

Solution

Step 1: Calculate the number of moles of each gas present in the container. Step

2: Use the ideal gas law to determine the partial pressure of each gas. Step 3:

Calculate the total pressure in the container.

Step 1: Calculate the number of moles of each gas present in the container.

1. Calculate the number of moles of nitrogen gas (N2):

moles of N2=mass

molar mass =1.00 g

28.0 g/mol = 0.0357 mol

2. Calculate the number of moles of oxygen gas (O2):

moles of O2=mass

molar mass =1.50 g

32.0 g/mol = 0.0469 mol

Step 2: Use the ideal gas law to determine the partial pressure of each gas.

The ideal gas law is given by:

P V =nRT

where: P= pressure (in atm) V= volume (in L) n= number of moles R=

universal gas constant T= temperature (in K)

1. For nitrogen gas:

PN2=n·R·T

V=0.0357 mol ·0.0821 L atm mol−1K−1·(25 + 273) K

2.00 L = 0.466 atm

2. For oxygen gas:

PO2=n·R·T

V=0.0469 mol ·0.0821 L atm mol−1K−1·(25 + 273) K

2.00 L = 0.612 atm

Step 3: Calculate the total pressure in the container. Total pressure =

PN2+PO2Total pressure = 0.466 atm + 0.612 atm Total pressure = 1.08 atm

Therefore, the partial pressure of nitrogen gas is 0.466 atm, the partial pres-

sure of oxygen gas is 0.612 atm, and the total pressure in the container is 1.08

atm.

Question 23

Question

When 10.0 grams of ethanol (C2H5OH) is burned in oxygen gas according to

the following reaction:

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)

how many moles of oxygen gas are required?

Solution

Step 1: Calculate the molar mass of ethanol. The molar mass of C2H5OH can

be calculated by summing the molar masses of its constituent atoms:

2×Atomic mass of C + 6 ×Atomic mass of H + 1 ×Atomic mass of O

Step 2: Convert the mass of ethanol from grams to moles. Use the molar

mass of ethanol calculated in step 1 to convert grams of ethanol to moles.

Step 3: Determine the mole ratio between ethanol and oxygen. From the

balanced chemical equation, we can see that 1 mole of ethanol requires 3 moles

of oxygen.

Step 4: Calculate the moles of oxygen required. Multiply the moles of

ethanol by the mole ratio of oxygen to ethanol.

Step 1:

The molar mass of ethanol (C2H5OH) is:

2×12.01 g/mol + 6 ×1.01 g/mol + 1 ×16.00 g/mol = 46.08 g/mol

Step 2:

Converting the mass of ethanol to moles:

Moles of C2H5OH =10.0 g

46.08 g/mol ≈0.217 moles

Step 3:

The mole ratio between ethanol and oxygen is 1:3. This means that 1 mole of

ethanol reacts with 3 moles of oxygen.

Step 4:

Calculating the moles of oxygen required:

Moles of O2= 0.217 moles ×3 moles O2

1 mole C2H5OH = 0.651 moles

Therefore, 0.651 moles of oxygen gas are required to burn 10.0 grams of

ethanol.

Question 24

Question

A mixture of 3.00 L of oxygen gas at STP and 5.00 L of hydrogen gas at 27

and 1.20 atm is allowed to react. The reaction is given by the equation:

2 H2(g) + O2(g)→2 H2O(g)

Calculate the total volume of the reaction vessel after the reaction is completed,

assuming the reaction goes to completion.

Solution

Step 1: Calculate the moles of oxygen gas present. Given:

Volume of oxygen gas (V) = 3.00 L

Pressure of oxygen gas (P) = 1.00 atm

Temperature of oxygen gas (T) = 273 K

Using the ideal gas law P V =nRT , we can calculate the moles of oxygen

gas:

n=P V

=(1.00 atm)(3.00 L)

0.0821 atm ·L/mol ·K·273 K

≈0.122 moles

Step 2: Calculate the moles of hydrogen gas present. Given:

Volume of hydrogen gas (V) = 5.00 L

Pressure of hydrogen gas (P) = 1.20 atm

Temperature of hydrogen gas (T) = 27C= 300 K

Using the ideal gas law, we can calculate the moles of hydrogen gas:

n=P V

=(1.20 atm)(5.00 L)

0.0821 atm ·L/mol ·K·300 K

≈0.243 moles

Step 3: Determine the limiting reactant. From the balanced chemical equa-

tion, we see that 1 mole of O2 reacts with 2 moles of H2. Therefore, the limiting

reactant is the one that produces the least amount of water, which in this case

is oxygen.

Step 4: Calculate the moles of water produced. Since oxygen is the limiting

reactant, all of it will react with hydrogen to produce water. The reaction

consumes 0.122 moles of oxygen, producing 0.244 moles of water.

Step 5: Calculate the volume of water vapor formed. Based on the stoichiom-

etry of the reaction, every 2 moles of water vapor occupies the same volume as 1

mole of oxygen. Thus, the total volume of water vapor formed can be calculated

as:

Volume of water vapor = 2

1×Volume of oxygen

= 2 ×3.00 L

= 6.00 L

Therefore, the total volume of the reaction vessel after the reaction is com-

pleted is 6.00 L.

Question 25

Question

A gaseous compound containing only hydrogen and nitrogen is analyzed and

found to be 85.7

(Given: molar volume at STP = 22.4 L/mol, R= 0.0821 atm ·L/mol ·K)

Solution

Step 1: Calculate the molar mass of the compound using the percent composi-

tion of nitrogen.

molar mass of N = 14.01 g/mol

molar mass of H = 1.01 g/mol

molar mass of compound = 100 g/mol = 85.7

14.01 +14.3

1.01

= 684.51 g/mol

Step 2: Use the ideal gas law to find the number of moles of the compound

present in the sample.

P V =nRT

n=P V

RT =(755 mmHg ×0.825 L)

(0.0821 atm ·L/mol ·K×298 K)

= 0.2780 mol

Step 3: Calculate the number of moles of nitrogen and hydrogen in the

compound and determine the empirical formula.

moles of N = 0.857 ×0.2780 = 0.2381 mol

moles of H = (1 −0.857) ×0.2780 = 0.0399 mol

Dividing by the smallest value gives the empirical formula NH.

Step 4: Determine the molecular formula using the molar mass calculated

in Step 1.

molar mass of empirical formula (NH) = 68 g/mol

molecular formula = 684.51

68 = 10

Therefore, the molecular formula of the compound is NH.

Question 26

Question

A 2.00 L container is filled with 5.00 mol of helium gas (He) at 25

C. If the

container is then heated to 80

C, what will be the new pressure inside the

container? (Assume ideal gas behavior)

Solution

Step 1: Calculate the initial pressure inside the container using the ideal gas

law:

P V =nRT

where: P= pressure (in atm) V= volume (in L) n= moles of gas R= ideal

gas constant = 0.0821 atm L / mol K T= temperature (in K)

Given: V= 2.00 L, n= 5.00 mol, T= 25

C = 298 K

Substitute these values into the ideal gas law:

P1=nRT1

P1=(5.00 mol)(0.0821 atm L/mol K)(298 K)

2.00 L

P1= 61.19 atm

Step 2: Calculate the final pressure inside the container using the new tem-

perature: Given: new temperature T2= 80

C = 353 K

Since the gas is kept at constant volume, we can use the following formula:

=P2

Substitute the initial pressure and temperature, and solve for the final pres-

sure P2:61.19

298 =P2

353

P2= 69.85 atm

Therefore, the new pressure inside the container when heated to 80

C will

be 69.85 atm.

Question 27

Question

When 4.50 L of ethylene gas (C2H4) reacts with excess oxygen gas in a com-

bustion reaction, 8.76 L of carbon dioxide gas (CO2) is produced. If all gas

volumes are measured at the same temperature and pressure, what is the bal-

anced chemical equation for the reaction?

Solution

Step 1: Write the balanced chemical equation for the reaction. Step 2: Use the

ideal gas law to find the number of moles of each reactant and product. Step

3: Determine the stoichiometry of the reaction based on the mole ratio of the

reactants and products. Step 4: Write the balanced chemical equation.

Step 1: Write the balanced chemical equation for the reaction.

Let the balanced chemical equation for the reaction be:

aC2H4+bO2→cCO2+dH2O

Step 2: Use the ideal gas law to find the number of moles of each reactant

and product.

Given: Volume of C2H4= 4.50 L Volume of CO2produced = 8.76 L

Since all gases are measured at the same temperature and pressure, their

volumes are directly proportional to the number of moles.

V olume1

V olume2

=Moles1

Moles2

Therefore, the number of moles of CO2produced is twice the number of

moles of C2H4consumed.

Step 3: Determine the stoichiometry of the reaction based on the mole ratio

of the reactants and products.

From the balanced chemical equation: - The coefficient in front of C2H4is

1. - The coefficient in front of CO2is c (which we don’t know yet).

This means that the coefficient in front of CO2should be 2 in order to

balance the equation.

Therefore, the balanced chemical equation is:

C2H4+ 3O2→2CO2+ 2H2O

Step 4: Write the balanced chemical equation.

The balanced chemical equation for the reaction is:

C2H4+ 3O2→2CO2+ 2H2O

Question 28

Question

A gaseous hydrocarbon is burned completely in oxygen gas to produce carbon

dioxide and water. If 0.85 moles of carbon dioxide are produced, how many

moles of the hydrocarbon were burned?

Solution

Step 1: Write the balanced chemical equation for the combustion of the hydro-

carbon.

Hydrocarbon + O2→CO2+ H2O

Step 2: Determine the molar ratio between carbon dioxide and the hydro-

carbon from the balanced chemical equation. From the equation, we see that

one mole of hydrocarbon produces one mole of carbon dioxide.

Step 3: Use the molar ratio to find the number of moles of the hydrocarbon

burned. Since 0.85 moles of carbon dioxide were produced, then 0.85 moles of

the hydrocarbon were burned as well.

Answer: 0.85 moles of the hydrocarbon were burned.

Question 29

Question

A mixture of 0.500 mol of methane (CH4) and 0.800 mol of oxygen (O2) is

placed in a container and ignited to produce carbon dioxide (CO2) and water

(H2O) according to the following unbalanced equation:

CH4+ O2→CO2+ H2O

Calculate the maximum amount of water (in grams) that can be produced from

this reaction.

Solution

Step 1: Write the balanced chemical equation for the reaction.

CH4+ 2O2→CO2+ 2H2O

Step 2: Determine the limiting reactant.

Given the mixture contains 0.500 mol of CH4and 0.800 mol of O2, we can

calculate the theoretical yield of the products for each reactant separately.

For CH4: - 0.500 mol of CH4will produce 2 * 0.500 mol = 1.000 mol of H2O

For O2: - 0.800 mol of O2will produce 2 * 0.800 mol = 1.600 mol of H2O

Since CH4produces a lower amount of water (1.000 mol), it is the limiting

reactant.

Step 3: Calculate the maximum amount of water produced.

The molar mass of water (H2O) is 18.015 g/mol.

Mass of water = 1.000 mol ×18.015 g/mol = 18.015 g

Therefore, the maximum amount of water that can be produced from this

reaction is 18.015 grams.

Question 30

Question

A gaseous hydrocarbon compound contains only carbon and hydrogen. When

6.80 g of this compound is burned in excess oxygen, 21.6 g of carbon dioxide

is produced. Determine the molecular formula of the compound assuming it is

composed of only carbon and hydrogen.

Solution

Step 1: Write the balanced chemical equation for the combustion of the hydro-

carbon compound:

Hydrocarbon + O2→CO2+ H2O

Step 2: Calculate the moles of carbon dioxide produced: Given mass of car-

bon dioxide = 21.6 g Molar mass of carbon dioxide (CO2) = 12.01 + 2(16.00) =

44.01 g/mol

Number of moles of carbon dioxide = 21.6 g

44.01 g/mol ≈0.491 mol

Step 3: Determine the moles of carbon in the hydrocarbon: 1 mol of hydro-

carbon (CxHy) produces 1 mol of CO2(from the balanced equation) So, the

number of moles of carbon in the hydrocarbon = 0.491 mol

Step 4: Calculate the mass of carbon in the hydrocarbon: Molar mass of

carbon (C) = 12.01 g/mol

Mass of carbon in 6.80 g of the hydrocarbon = 6.80 g×0.491 mol

1 mol ×12.01 g/mol ≈

39.44 g

Step 5: Determine the mass of hydrogen in the hydrocarbon: Mass of hy-

drogen in the hydrocarbon = Total mass of the hydrocarbon - Mass of carbon

Mass of hydrogen = 6.80 g - 39.44 g = 32.36 g

Step 6: Calculate the moles of hydrogen in the hydrocarbon: Molar mass of

hydrogen (H) = 1.01 g/mol

Number of moles of hydrogen = 32.36 g

1.01 g/mol ≈32.04 mol

Step 7: Find the ratio of moles of carbon to hydrogen: Divide the moles of

carbon by the moles of hydrogen: 0.491 mol

32.04 mol ≈0.0153

Step 8: Determine the empirical formula of the hydrocarbon compound: The

empirical formula of the compound is CH(1

0.0153 )≈CH65

Step 9: Calculate the molecular formula of the hydrocarbon: Given the

molecular formula mass, MM, is 21.6 g ⇒MMCH65 = 12.01 + 1(1.01) =

13.02 g/mol

The molecular formula is thus CH65 ×21.6 g/mol

13.02 g/mol ≈CH111

Question 31

Question

A reaction occurs between 5.00 L of propane gas (C3H8) and 9.00 L of oxygen

gas (O2) at a certain temperature and pressure. The reaction forms carbon

dioxide (CO2) and water vapor (H2O). If all of the propane is consumed in the

reaction, what volume of carbon dioxide is produced? Assume all gases are at

the same temperature and pressure.

Solution

Step 1: Write the balanced chemical equation for the reaction between propane

and oxygen. The balanced chemical equation for the complete combustion of

propane is:

C3H8+ 5O2−→ 3CO2+ 4H2O

Step 2: Determine the limiting reactant. To determine the limiting reactant,

we need to first calculate the number of moles of each reactant. Given: Volume

of C3H8= 5.00 L Volume of O2= 9.00 L

Using the ideal gas law, we can convert the volume of gases to moles:

n=P V

Assuming temperature and pressure are constant, we get:

nC3H8=P V

RT =(5)(V)

nO2=P V

RT =(9)(V)

Step 3: Determine the limiting reactant. The stoichiometry of the reaction

indicates that 1 mol of C3H8reacts with 5 mol of O2to produce 3 mol of CO2.

Calculate the moles of C3H8and O2:

Moles of C3H8=5

Moles of O2=9

Since the reaction requires 5 moles of O2for every mole of C3H8,O2is the

limiting reactant.

Step 4: Calculate the volume of CO2produced. From the balanced chemical

equation, 1 mole of C3H8produces 3 moles of CO2. Since O2is the limiting

reactant, we can determine the moles of CO2produced using the stoichiometry:

Moles of CO2= (3)( 9

RT ) = 27

Now, we can calculate the volume of CO2using the ideal gas law:

VCO2=27V

Question 32

Question

A 2.00 L flask contains 1.00 atm of N2, 2.00 atm of H2, and 3.00 atm of NH3

at equilibrium in the reaction:

N2(g) + 3H2(g)⇌2NH3(g)

If the temperature is held constant, will the pressure change if an additional

0.500 atm of H2is added to the flask? Justify your answer.

Solution

Step 1: Write down the balanced chemical equation for the reaction.

The balanced chemical equation is:

N2(g) + 3H2(g)⇌2NH3(g)

Step 2: Determine the initial moles and partial pressures of the gases.

Using the ideal gas law P V =nRT , we can calculate the initial moles of gas

in the flask as follows:

For N2:

n(N2) = P V

RT =(1.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

For H2:

n(H2) = P V

RT =(2.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

For NH3:

n(NH3) = P V

RT =(3.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

Step 3: Check the reaction quotient Q and compare it to the equilibrium

constant K.

The reaction quotient Q is given by:

Q=[NH3]2

[N2][H2]3

If Q = K, the system is at equilibrium; if Q ¿ K, the reaction shifts left; if

Q ¡ K, the reaction shifts right.

Step 4: Determine the final moles and partial pressures of the gases.

When 0.500 atm of H2is added to the flask, the partial pressure of H2will

increase. This will cause Q to become greater than K, prompting the reaction

to shift to the left in order to re-establish equilibrium. This shift will result in

changes to the moles and partial pressures of the gases in the flask.

Question 33

Question

A reaction occurs between 12.0 L of hydrogen gas at STP and excess nitrogen

gas to produce ammonia gas. If the reaction is 85

Solution

Step 1: Write the balanced chemical equation for the reaction between hydrogen

and nitrogen to produce ammonia.

3H2(g)+N2(g)→2NH3(g)

Step 2: Determine the moles of hydrogen gas present using the ideal gas law,

P V =nRT .

PV = nRT

(1 atm)(12.0 L) = n ×(0.08206 L atm / mol K) ×(273 K)

n = (1)(12.0)

(0.08206)(273) = 0.532 mol

Step 3: Calculate the theoretical yield of ammonia gas based on the number

of moles of hydrogen gas used (using the stoichiometry of the balanced equation).

Moles of NH3=(0.85)(0.532)

3= 0.150 mol

Step 4: Convert the moles of ammonia gas to volume using the ideal gas

law.

PV = nRT

V = nRT

P=(0.150)(0.08206)(273)

1= 3.958 L

Therefore, the volume of ammonia gas that can be produced is 3.958 L.

Question 34

Question

Magnesium reacts with hydrochloric acid to produce magnesium chloride and

hydrogen gas. If 3.00 g of magnesium is reacted with excess hydrochloric acid

to produce 0.500 L of hydrogen gas at 298 K and 1.00 atm, what is the percent

yield of this reaction? (Assume the reaction goes to completion.)

Solution

Step 1: Write the balanced chemical equation for the reaction:

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the moles of magnesium used: Given: mass = 3.00 g,

molar mass of Mg = 24.31 g/mol

moles of Mg = 3.00 g

24.31 g/mol = 0.1235 mol

Step 3: Determine the moles of hydrogen gas produced using stoichiometry:

From the balanced equation, 1 mole of Mg produces 1 mole of H2 gas.

moles of H2= 0.1235 mol

Step 4: Calculate the volume of hydrogen gas produced using the ideal gas

law: Given: volume = 0.500 L, temperature = 298 K, pressure = 1.00 atm,

R=0.0821 L ·atm/K·mol

moles of H2=pressure ×volume

R×temperature =1.00 atm ×0.500 L

0.0821 L ·atm/K·mol ×298 K ≈0.0201 mol

Step 5: Calculate the theoretical yield of hydrogen gas: Theoretical yield is

the lower of the two quantities found in steps 3 and 4.

theoretical yield = 0.0201 mol

Step 6: Calculate the percent yield: Given: actual yield = 0.0201 mol,

theoretical yield = 0.0201 mol

percent yield = actual yield

theoretical yield×100% = 0.0201 mol

0.0201 mol×100% = 100%

Therefore, the percent yield of this reaction is 100

XO2= 1 −0.600

XO2= 0.400

Therefore, the moles of nitrogen gas (nN2) and oxygen gas (nO2) can be

calculated as:

nN2=XN2×n

nN2= 0.600 ×0.0452

nN2= 0.0271 moles

nO2=XO2×n

nO2= 0.400 ×0.0452

nO2= 0.0181 moles

Step 3: Calculate the partial pressure of oxygen gas. The partial pressure

of oxygen gas (PO2) can be calculated using the formula:

PO2=nO2RT

PO2=(0.0181 moles)(0.0821 L atm/mol K)(300 K)

1.50 L

Calculating PO2, we find:

PO2≈0.0915 atm

Therefore, the partial pressure of oxygen gas in the container is approxi-

mately 0.0915 atm.

Question 2

Question

A gaseous compound containing only carbon and hydrogen is burned in air. A

1.50 g sample of the compound yields 4.20 g of carbon dioxide and 1.71 g of

water. Determine the empirical formula of the compound.

Solution

Step 1: Calculate the moles of carbon dioxide produced. Given: Mass of carbon

dioxide produced = 4.20 g

Molar mass of carbon dioxide (CO2) = 44.01 g/mol

Number of moles of CO2=4.20 g

44.01 g/mol = 0.0954 mol

Step 2: Calculate the moles of water produced. Given: Mass of water pro-

duced = 1.71 g

Molar mass of water (H2O) = 18.02 g/mol

Number of moles of H2O = 1.71 g

18.02 g/mol = 0.0949 mol

Step 3: Determine the moles of carbon and hydrogen in the compound.

From the balanced chemical equation for the combustion of the compound: 1

mol of the compound yields 1 mol of CO2and 1 mol of H2O

Therefore, the moles of carbon in the compound = 0.0954 mol The moles of

hydrogen in the compound = 0.0949 mol

Step 4: Calculate the mole ratio of carbon to hydrogen. Assume the com-

pound has the general formula CaHb, where ’a’ and ’b’ are the subscripts we

want to determine.

From the moles of carbon and hydrogen: a=0.0954

0.0949 ≈1

b=0.0949

0.0949 = 1

Thus, the empirical formula of the compound is CH.

Step 5: Write the empirical formula. The empirical formula of the compound

is CH.

Question 3

Question

A 1.00 L container holds a mixture of oxygen gas and nitrogen gas at a total

pressure of 3.00 atm and a temperature of 25

C. The partial pressure of oxygen

gas is 1.20 atm. If all the oxygen gas is consumed in a reaction that produces

nitrogen dioxide gas, what will be the partial pressure of nitrogen dioxide gas in

the container at the end of the reaction? Assume the reaction goes to completion

and the volume and temperature remain constant.

Solution

Step 1: Calculate the initial moles of oxygen gas using the ideal gas law:

n=P V

Given:

Poxygen = 1.20 atm

Ptotal = 3.00 atm

V= 1.00 L

T= 25C= 298 K

R= 0.0821 L

atm/mol

Plugging in the values:

noxygen =(1.20 atm)(1.00 L)

(0.0821 L

atm/mol

K)(298 K)

noxygen ≈0.0589 mol

Step 2: Since the reaction goes to completion without change in volume or

temperature, all the oxygen gas reacts to form nitrogen dioxide gas. Therefore,

the moles of nitrogen dioxide gas produced will be equal to the moles of oxygen

consumed.

Step 3: Calculate the mole fraction of nitrogen dioxide gas:

x=nnitrogen dioxide

ntotal

Since the total moles of gas is the sum of moles of oxygen and moles of

nitrogen dioxide:

ntotal =noxygen +nnitrogen dioxide

Given that the moles of nitrogen dioxide gas produced is equal to the moles

of oxygen consumed:

ntotal = 2noxygen

Substitute the values:

ntotal = 2(0.0589)

ntotal = 0.118 mol

Step 4: Calculate the total pressure of the mixture after the reaction:

Ptotal, final =Pnitrogen dioxide +Pnitrogen

Given that the moles of nitrogen gas remains unchanged:

Ptotal, final =Pnitrogen dioxide +Pnitrogen = (nnitrogen dioxide +nnitrogen)RT/V

Since the partial pressure of nitrogen gas is the same as the initial condition:

Pnitrogen = (nnitrogen)RT/V = (0.118 mol)0.0821 L

atm/mol

K(298 K)/1.00 L

Now, calculate the partial pressure of nitrogen dioxide gas:

Pnitrogen dioxide =Ptotal, final −Pnitrogen

Pnitrogen dioxide = 3.00 atm−(0.118 mol ×0.0821 L

atm/mol

K×298 K/1.00 L)

Pnitrogen dioxide = 3.00 atm −0.287 atm

Pnitrogen dioxide = 2.71 atm

Therefore, the partial pressure of nitrogen dioxide gas in the container at

the end of the reaction will be 2.71 atm.

Question 4

Question

A reaction occurs between 15.0 L of hydrogen gas at standard temperature and

pressure (STP) and excess oxygen gas. If the reaction produces water vapor,

what volume of water vapor is produced at STP?

Solution

Step 1: Write the balanced chemical equation for the reaction. The balanced

chemical equation for the reaction between hydrogen gas and oxygen gas to

produce water vapor is:

2H2(g) + O2(g)→2H2O(g)

Step 2: Determine the moles of hydrogen gas. Given that the volume of

hydrogen gas is 15.0 L at STP (standard temperature and pressure: 0

C and 1

atm), we can use the ideal gas law to find the number of moles of hydrogen gas:

P V =nRT

n=P V

RT =(1 atm)(15.0 L)

(0.0821 L ·atm/mol ·K)(273 K) = 0.681 moles

Step 3: Determine the volume of water vapor produced. From the balanced

chemical equation, we see that the ratio of moles of hydrogen gas to moles of

water vapor is 2:2. Therefore, the moles of water vapor produced will also be

0.681. Now, we can use this mole value to find the volume of water vapor

produced at STP:

P V =nRT

V=nRT

P=(0.681 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm = 15.2 L

Therefore, 15.2 L of water vapor is produced at STP.

Question 5

Question

A sample of iron metal reacts with sulfuric acid according to the following

balanced chemical equation:

Fe + H2SO4→FeSO4+ H2

If 5.00 grams of iron react with an excess of sulfuric acid, how many grams

of hydrogen gas are produced?

(Note: molar mass of iron = 55.85 g/mol, molar mass of sulfuric acid =

98.08 g/mol, molar mass of hydrogen gas = 2.02 g/mol)

Solution

Step 1: Calculate the number of moles of iron (Fe) given 5.00 grams of iron.

Moles of Fe = Mass

Molar Mass =5.00 g

55.85 g/mol ≈0.0894 mol

Step 2: Use the balanced chemical equation to determine the mole ratio be-

tween iron (Fe) and hydrogen gas (H¡sub¿2¡/sub¿). From the balanced chemical

equation, 1 mole of Fe produces 1 mole of H¡sub¿2¡/sub¿.

Step 3: Calculate the number of moles of hydrogen gas produced.

Moles of H2= Moles of Fe ×Mole Ratio = 0.0894 mol ×1 = 0.0894 mol

Step 4: Calculate the mass of hydrogen gas produced.

Mass of H2= Moles ×Molar Mass = 0.0894 mol ×2.02 g/mol = 0.181 g

Therefore, 0.181 grams of hydrogen gas are produced when 5.00 grams of

iron react with sulfuric acid.

Question 6

Question

A gaseous compound contains only carbon and hydrogen. When 0.500 mol of

the compound is reacted with excess oxygen, it produces 1.50 mol of carbon

dioxide and 2.00 mol of water vapor. Determine the molecular formula of the

compound.

Solution

Step 1: Write the balanced chemical equation for the reaction.

CxHy+ O2→CO2+ H2O

Step 2: Write the stoichiometry ratio from the given information. - 0.500 mol

of the compound produces 1.50 mol of carbon dioxide. This implies a 1:3 ratio

between the compound and carbon dioxide. - 0.500 mol of the compound pro-

duces 2.00 mol of water vapor. This implies a 1:4 ratio between the compound

and water vapor.

Step 3: Determine the molar masses of carbon dioxide (44 g/mol) and

water (18 g/mol). - The compound’s molar mass can be represented as x×

molar mass of C + y×molar mass of H.

Step 4: Determine the molar mass of the compound. - From the stoichiom-

etry ratios, we can construct a system of equations to find the values of x and

Step 5: Solve for x and y. - From the system of equations, we find that x =

4 and y = 10.

Step 6: Determine the molecular formula of the compound. - The compound

is C4H10, which is the molecular formula for butane.

Question 7

Question

A mixture of nitrogen gas (N2) and oxygen gas (O2) is held in a 5.00 L container

at a temperature of 300 K. The pressure of the mixture is measured to be 2.00

atm. If the mole fraction of nitrogen gas in the mixture is 0.60, calculate the

number of moles of oxygen gas in the container.

Solution

Step 1: Calculate the total moles of gas in the container.

Given that the mole fraction of nitrogen gas is 0.60, we can calculate the moles

of nitrogen gas in the container as follows:

Moles of N2= Mole fraction ×Total moles of gas

Moles of N2= 0.60 ×Total moles of gas

Step 2: Use the ideal gas law to find the total moles of gas.

The ideal gas law is given by:

P V =nRT

where: - P is the pressure in atm, - V is the volume in liters, - n is the number of

moles, - R is the ideal gas constant (0.0821 L·atm/mol·K), - T is the temperature

in Kelvin.

We can rearrange this equation to solve for n:

n=P V

Step 3: Substitute known values into the equation to find the total moles of

gas.

n=(2.00 atm) ×(5.00 L)

(0.0821 L·atm/mol·K) ×(300 K)

n≈10.00

24.63

n≈0.41 moles

Step 4: Calculate the moles of oxygen gas in the container.

Since the moles of nitrogen gas is 0.60 times the total moles of gas:

Moles of O2= (1 −0.60) ×Total moles of gas

Moles of O2= 0.40 ×0.41

Moles of O2= 0.164 moles

Therefore, there are 0.164 moles of oxygen gas in the 5.00 L container at 300

Question 8

Question

A reaction takes place between 2.50 L of hydrogen gas at STP and an excess of

oxygen gas. If the reaction yields water vapor as the only product, how many

grams of water vapor are produced?

Solution

Step 1: Determine the moles of hydrogen gas using the ideal gas law equation:

P V =nRT

Where: - P= pressure = 1 atm - V= volume = 2.50 L - n= moles - R= ideal

gas constant = 0.0821 L

atm/K

mol - T= temperature (STP) = 273 K

Substitute the values into the equation:

1 atm ×2.50 L = n×0.0821 L

atm/K

mol ×273 K

n=1×2.50

0.0821 ×273 ≈0.105 mol

Step 2: Using the balanced chemical equation for the reaction:

2 H2+ O2→2 H2O

1 mol of H2reacts to produce 2 mol of H2O.

Therefore, the moles of water vapor produced will be:

0.105 mol ×2 mol H2O

2 mol H2

= 0.105 mol H2O

Step 3: Calculate the molar mass of water (H2O):

2×1.01 g/mol (H) + 16.00 g/mol (O) = 18.02 g/mol

Step 4: Finally, determine the mass of water vapor produced:

0.105 mol H2O×18.02 g/mol = 1.89 g

Question 9

Question

A reaction between solid potassium chlorate and solid manganese dioxide pro-

duces potassium chloride, manganese(IV) oxide, and oxygen gas.

How many liters of oxygen gas at STP are produced when 25 g of potassium

chlorate reacts completely, assuming the reaction takes place at STP conditions?

Solution

Step 1: Write the balanced chemical equation for the reaction.

2KClO3(s)→2KCl(s) + 3O2(g)

Step 2: Calculate the molar mass of potassium chlorate (KClO).

K: 39.10 g/mol, Cl: 35.45 g/mol, O: 16.00 g/mol

Molar mass of KClO3= 39.10 + 35.45 + 3(16.00) = 122.55 g/mol

Step 3: Calculate the number of moles of potassium chlorate used.

Moles = Mass

Molar mass =25 g

122.55 g/mol = 0.204mol

Step 4: Use the balanced chemical equation to find the moles of oxygen gas

produced.

Moles of O2= 3 ×moles of KClO3= 3 ×0.204 mol = 0.612 mol

Step 5: Calculate the volume of oxygen gas at STP using the ideal gas law

(V=nRT

P).

R=0.0821 L·atm

K·mol,T = 273 K,P = 1 atm

V=0.612 mol ×0.0821 L·atm

K·mol ×273 K

1 atm = 13.7 L

So, 13.7 liters of oxygen gas at STP are produced when 25 g of potassium

chlorate reacts completely.

Question 10

Question

A reaction takes place between acetylene gas (C2H2) and oxygen gas according

to the following balanced chemical equation:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

If 10.0 L of acetylene gas at STP reacts completely with excess oxygen gas, what

volume of carbon dioxide is produced at STP?

Solution

Step 1: Find the number of moles of acetylene gas. Given that the volume of

acetylene gas is 10.0 L at STP, we can use the equation n=P V

RT to find the

number of moles.

For C2H2:nC2H2=(1.00 atm)(10.0 L)

(0.0821 L ·atm/K ·mol)(273 K) = 0.417 mol

Step 2: Determine the limiting reactant. From the balanced chemical equa-

tion, we can see that 2 moles of C2H2react with 5 moles of O2.

Moles of O2=5

2×moles of C2H2=5

2×0.417 = 1.04 mol

Step 3: Calculate the volume of carbon dioxide produced. From the balanced

chemical equation, we see that 2 moles of C2H2produce 4 moles of CO2.

Moles of CO2= 4 ×moles of C2H2= 4 ×0.417 = 1.67 mol

Now, we can calculate the volume of carbon dioxide produced using the ideal

gas law.

V=nRT

P=(1.67 mol)(0.0821 L ·atm/K ·mol)(273 K)

1.00 atm = 38.1 L

Therefore, 38.1 L of carbon dioxide is produced at STP.

Question 11

Question

A sample of sodium azide, NaN3, is heated and its decomposition reaction

produces nitrogen gas, sodium metal, and sodium oxide. If 5.00 L of nitrogen

gas is collected at 750 mmHg and 25

C, how many grams of sodium azide were

decomposed? (Assume the reaction goes to completion.)

Solution

Step 1: Write the balanced chemical equation for the decomposition of sodium

azide:

2NaN3→2Na + 3N2

Step 2: Calculate the number of moles of nitrogen gas collected using the

ideal gas law:

P V =nRT

n=P V

RT =(750 ×10−3atm) ×(5.00 L)

0.0821 L·atm

mol·K×(25 + 273) K

n= 0.0918 mol N2

Step 3: Use the balanced chemical equation to determine the moles of NaN3

decomposed: Since the molar ratio between NaN3and N2is 2:3, the moles of

NaN3decomposed is calculated as follows:

moles of NaN3=2

3×0.0918mol = 0.0612mol NaN3

Step 4: Calculate the molar mass of NaN3:

Molar mass of NaN3= 3(22.99) + 14.01 + 3(16.00) = 65.02 g/mol

Step 5: Determine the mass of sodium azide decomposed:

Mass of NaN3= 0.0612mol ×65.02 g/mol = 3.98 g

Therefore, 3.98 grams of sodium azide were decomposed.

Question 12

Question

A 2.50 L container at 27

C contains a mixture of 2.63 g of helium (He) and 8.65

g of neon (Ne) at a total pressure of 1.25 atm. Assuming ideal gas behavior,

determine the partial pressure of each gas in the mixture.

Solution

Step 1: Calculate the moles of helium and neon present. Step 2: Determine the

mole fraction of each gas. Step 3: Calculate the partial pressure of each gas

using the mole fractions.

Step 1: Calculate the moles of helium and neon present. The molar mass

of helium (He) is 4.00 g/mol, and the molar mass of neon (Ne) is 20.18 g/mol.

Number of moles of helium:

nHe =2.63 g

4.00 g/mol = 0.6575 mol

Number of moles of neon:

nNe =8.65 g

20.18 g/mol = 0.4281 mol

Step 2: Determine the mole fraction of each gas. Total moles of gas:

ntotal =nHe +nNe = 0.6575 mol + 0.4281 mol = 1.0856 mol

Mole fraction of helium:

XHe =nHe

ntotal

=0.6575 mol

1.0856 mol = 0.6060

Mole fraction of neon:

XNe =nNe

ntotal

=0.4281 mol

1.0856 mol = 0.3940

Step 3: Calculate the partial pressure of each gas using the mole fractions.

Partial pressure of helium:

PHe =XHe ·Ptotal = 0.6060 ×1.25 atm = 0.7575 atm

Partial pressure of neon:

PNe =XNe ·Ptotal = 0.3940 ×1.25 atm = 0.4925 atm

Therefore, the partial pressure of helium is 0.7575 atm and the partial pres-

sure of neon is 0.4925 atm in the mixture.

Question 13

Question

A gaseous compound contains only carbon, hydrogen, and oxygen. When a

0.500 g sample of the compound is completely combusted in excess oxygen,

1.100 g of carbon dioxide and 0.450 g of water are produced. Determine the

empirical formula of the compound.

(Given: molar mass of carbon dioxide = 44.01 g/mol, molar mass of water

= 18.015 g/mol)

Solution

Step 1: Calculate the moles of carbon dioxide produced. Let x be the moles of

carbon in the compound.

Moles of CO2=1.100 g

44.01 g/mol = 0.025 mol

Step 2: Calculate the moles of water produced. From the given information,

we can calculate the moles of hydrogen in the compound. Moles of hydrogen in

water = 0.450 g

18.015 g/mol = 0.025 mol

Step 3: Calculate the moles of oxygen in the compound. Let y be the moles

of oxygen in the compound. From the combustion reaction, the moles of oxygen

can be determined based on the moles of carbon and hydrogen. The moles of

oxygen can be calculated using the mole ratios in the compound.

Step 4: Determine the empirical formula of the compound. The ratio of

moles of carbon, hydrogen, and oxygen in the compound can be used to write

the empirical formula.

Thus, the empirical formula of the compound is determined to be C1H1O1,

which simplifies to CH2O.

Question 14

Question

A mixture of methane gas (CH4) and oxygen gas (O2) is combusted in a closed

container. If 15.0 g of methane reacts with excess oxygen to produce carbon

dioxide and water, determine the volume of carbon dioxide produced at STP.

Assume complete combustion.

Solution

Step 1: Write the balanced chemical equation for the combustion of methane.

CH4+ 2O2→CO2+ 2H2O

Step 2: Calculate the number of moles of methane used in the reaction.

Given: Mass of methane = 15.0 g Molar mass of methane (CH4) = 16.04

g/mol

Number of moles of methane = 15.0 g

16.04 g/mol = 0.9351 mol

Step 3: Determine the number of moles of carbon dioxide produced using

the stoichiometry of the balanced chemical equation.

From the balanced equation, 1 mol of methane produces 1 mol of carbon

dioxide. Number of moles of carbon dioxide = 0.9351 mol

Step 4: Calculate the volume of carbon dioxide produced at STP.

Given: Standard Temperature and Pressure (STP) conditions are 0

C (273 K)

and 1 atm pressure. Molar volume of gas at STP = 22.4 L/mol

Volume of carbon dioxide = 0.9351 mol ×22.4 L/mol = 20.95 L

Answer: The volume of carbon dioxide produced at STP is 20.95 L.

Question 15

Question

Methane (CH4) undergoes combustion according to the following balanced

chemical equation:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

If 8.00 L of methane is reacted with excess oxygen, how many moles of water

vapor will be produced? Assume all gases are at the same temperature and

pressure.

Solution

Step 1: Write down the given information and the balanced chemical equation.

Given: Volume of CH4= 8.00LCoefficients of H2O(g) in the balanced chemical

equation = 2

Balanced chemical equation:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Step 2: Find the number of moles of CH4. Given volume of CH4= 8.00L

Using the ideal gas law, we can convert the volume to moles:

n=P V

Step 3: Calculate the number of moles of H2Oproduced. From the balanced

chemical equation, we see that 1 mole of CH4produces 2 moles of H2O.

So, moles of H2Oproduced = moles of CH4×2

Step 4: Substitute the values into the equations and solve for the number of

moles of H2Oproduced.

n=P V

nCH4=(1)(8.00)

(0.0821)(298)

nCH4≈0.326 moles

Moles of H2Oproduced = 0.326 ×2=0.652 moles

Therefore, 0.652 moles of water vapor will be produced when 8.00 L of

methane undergoes combustion.

Question 16

Question

A reaction between solid magnesium and hydrochloric acid produces hydrogen

gas. If 3.00 g of magnesium reacts completely with an excess of hydrochloric

acid to produce 2.15 L of hydrogen gas at a temperature of 25

C and a pressure

of 1.00 atm, determine the balanced chemical equation for the reaction and

the volume of hydrogen gas at STP that would be produced from 4.50 g of

magnesium.

Solution

Step 1: Determine the balanced chemical equation for the reaction between

magnesium and hydrochloric acid.

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the number of moles of hydrogen gas produced in the

reaction with 3.00 g of magnesium.

Moles of Mg = 3.00 g

24.305 g/mol = 0.1235 mol

Moles of H2= 0.1235 mol ×1 mol H2

1 mol Mg = 0.1235 mol

Step 3: Use the ideal gas law to determine the volume of hydrogen gas

produced at STP.

PV = nRT

(1.00 atm)(2.15 L) = (0.1235 mol)(0.0821 L ·atm/mol ·K)(298 K)

Volume of H2=0.1235 ×0.0821 ×298 ×2.15

1.00 ≈6.69 L at STP

Step 4: Determine the volume of hydrogen gas produced at STP from 4.50

g of magnesium.

Moles of Mg = 4.50 g

24.305 g/mol = 0.1853 mol

Moles of H2= 0.1853 mol ×1 mol H2

1 mol Mg = 0.1853 mol

Volume of H2=0.1853 ×0.0821 ×298 ×2.15

1.00 ≈8.23 L at STP

Therefore, the volume of hydrogen gas produced at STP from 4.50 g of

magnesium is approximately 8.23 L.

Question 17

Question

A 2.00 L container at 25

C is filled with 0.400 mol of nitrogen gas and 1.00

mol of oxygen gas. If the two gases react according to the following balanced

equation:

N2(g)+O2(g)→2NO(g)

What is the total pressure in the container at 25

C after the reaction is complete?

Assume ideal gas behavior.

Solution

Step 1: Write down the balanced chemical equation.

N2(g)+O2(g)→2NO(g)

Step 2: Calculate the moles of nitrogen and oxygen gas initially. - Moles of

nitrogen gas, nN2= 0.400 mol - Moles of oxygen gas, nO2= 1.00 mol

Step 3: Determine the limiting reactant by comparing the mole ratios in the

balanced equation. Since the stoichiometry of the balanced equation is 1:1 for

N2and O2, oxygen gas is the limiting reactant.

Step 4: Use the ideal gas law to find the partial pressure of each gas before

the reaction. - For nitrogen gas: PN2=nN2RT

V=(0.400mol)(0.0821 atm·L

mol·K)(298K)

2.00L=

4.94atm - For oxygen gas: PO2=nO2RT

V=(1.00mol)(0.0821 atm·L

mol·K)(298K)

2.00L= 12.4atm

Step 5: Determine the change in moles for the gases. - For nitrogen gas:

∆nN2=−1(0.400) = −0.400 mol - For oxygen gas: ∆nO2=−1(1.00) = −1.00

mol - For nitrogen monoxide gas: ∆nN O = +2(1.00) = 2.00 mol

Step 6: Calculate the total moles of gas at the end. - Total moles of gas

=nN2+nO2+ ∆nNO = 0.400 + 1.00 + 2.00 = 3.40 mol

Step 7: Use the ideal gas law again to find the total pressure in the container

at the end.

Ptotal =ntotalRT

V=(3.40mol)(0.0821 atm·L

mol·K)(298K)

2.00L= 14.2atm

Therefore, the total pressure in the container at 25

C after the reaction is

complete is 14.2 atm.

Question 18

Question

A 2.50 L container at 25

C and 1.00 atm contains 0.500 mol of nitrogen gas,

1.00 mol of oxygen gas, and 2.00 mol of argon gas. If the container is heated to

160

C, what will be the new pressure inside the container?

Solution

Step 1: Calculate the initial total pressure inside the container using the ideal

gas law:

P V =nRT

Given: Initial volume (V) = 2.50 L, Initial temperature (T) = 25C= 25 +

273.15 K, Initial pressure (P)=1.00 atm, Moles of nitrogen (nN2)=0.500 mol,

Moles of oxygen (nO2) = 1.00 mol, Moles of argon (nAr)=2.00 mol.

First, let’s find the total moles of gas in the container:

ntotal =nN2+nO2+nAr

ntotal = 0.500 + 1.00 + 2.00 = 3.50 mol

Now, substitute the given values into the ideal gas law to find the initial

pressure:

Pinitial =ntotalRT

Pinitial =3.50 ×0.0821 ×(25 + 273.15)

2.50

Pinitial =3.50 ×0.0821 ×298.15

2.50

Pinitial =86.487

2.50

Pinitial = 34.59 atm

Step 2: Calculate the final pressure inside the container after heating the

gases to 160

C using the ideal gas law:

Pfinal =ntotalRT

Substitute the new temperature (T= 160 + 273.15 K) into the ideal gas law

to find the final pressure:

Pfinal =3.50 ×0.0821 ×(160 + 273.15)

2.50

Pfinal =3.50 ×0.0821 ×433.15

2.50

Pfinal =123.929

2.50

Pfinal = 49.57 atm

Therefore, the new pressure inside the container after heating the gases to

160

C will be 49.57 atm.

Question 19

Question

A 2.00 L container holds a mixture of H2, O2, and H2O gases at 425◦C and

1.00 atm. If the total pressure in the container is 3.00 atm, what is the partial

pressure of each gas?

Solution

Step 1: Calculate the moles of each gas based on the ideal gas law, P V =nRT ,

where Pis the pressure, Vis the volume, nis the number of moles, Ris the

gas constant, and Tis the temperature in Kelvin.

Given: P= 1.00 atm = 1.013 ×105Pa V= 2.00 L T= 425◦C = (425 +273)

K = 698 K R= 8.314 J/(mol·K)

For H2: 1.00 ×2.00 = n×8.314 ×698 n=1.00×2.00

8.314×698

Step 2: Repeat the calculation for O2and H2O.

nO2 =2.00×2.00

8.314×698 nH2O =(3.00−1.00)×2.00

8.314×698

Step 3: Calculate the mole fractions of each gas.

Xi=ni

ntotal

Step 4: Calculate the partial pressure of each gas using Dalton’s law of

partial pressures, Pi=Xi·Ptotal.

PH2 =XH2 ×1.00 PO2 =XO2 ×1.00 PH2O =XH2O ×2.00

Question 20

Question

A gas mixture contains helium and argon. If 8.00 L of the mixture at 27

C and

1.00 atm pressure contains 0.500 mol of helium and 1.00 mol of argon, what is

the partial pressure of each gas in the mixture?

Solution

Step 1: Calculate the total number of moles in the mixture. Let ntotal be the

total number of moles in the mixture. We are given that there are 0.500 mol of

helium and 1.00 mol of argon.

ntotal = 0.500 mol + 1.00 mol = 1.50 mol

Step 2: Calculate the mole fraction of each gas. The mole fraction of helium,

denoted by XHe, is defined as the moles of helium divided by the total moles in

the mixture. Similarly, the mole fraction of argon, denoted by XAr, is given by

the moles of argon divided by the total moles in the mixture.

XHe =0.500 mol

1.50 mol =1

XAr =1.00 mol

1.50 mol =2

Step 3: Calculate the partial pressure of each gas. The partial pressure of a

gas in a mixture is given by the product of the mole fraction of the gas and the

total pressure of the mixture.

PHe =XHe ×Ptotal =1

3×1.00 atm = 1

3atm

PAr =XAr ×Ptotal =2

3×1.00 atm = 2

3atm

Therefore, the partial pressure of helium is 1

3atm and the partial pressure

of argon is 2

3atm in the mixture.

Question 21

Question

A chemical reaction produces 8.60 L of carbon dioxide gas (CO2) at a temper-

ature of 25

C and pressure of 2.00 atm. If the reaction is known to produce

1.5 moles of CO2, what is the volume of oxygen gas (O2) produced at the same

conditions?

(Note: Assume all gases are ideal and use standard temperature and pressure

(STP).)

Solution

Step 1: Calculate the number of moles of CO2produced using the ideal gas law

formula:

PV = nRT

Given that the pressure (P) = 2.00 atm, volume (V) = 8.60 L, temperature

(T) = 25

C (which is 298 K), and we are solving for the number of moles (n):

nCO2=PV

RT =(2.00 atm)(8.60 L)

0.0821 L atm/mol K ×298 K

nCO2= 0.718 mol

Step 2: Use the stoichiometry of the reaction to find the number of moles

of O2produced. Since we know that 1.5 moles of CO2produces x moles of O2,

we can set up the following mole ratio:

nO2

nCO2

=1.5 mol CO2

1 mol O2

Substitute in the value of nCO2from step 1:

nO2

0.718 mol CO2

=1.5 mol CO2

1 mol O2

nO2= 1.078 mol O2

Step 3: Calculate the volume of O2gas produced using the ideal gas law

formula with the same temperature and pressure:

PV = nRT

VO2=nO2×RT

P=(1.078 mol)(0.0821 L atm/mol K ×298 K)

2.00 atm

VO2= 13.2 L

Therefore, the volume of O2gas produced at the same conditions is 13.2 L.

Question 22

Question

A 2.00 L container holds 1.00 g of nitrogen gas and 1.50 g of oxygen gas at 25

What is the partial pressure of each gas, and what is the total pressure in the

container?

Given: Molar mass of N2= 28.0 g/mol Molar mass of O2= 32.0 g/mol

Universal gas constant, R = 0.0821 L atm mol−1K−1

Solution

Step 1: Calculate the number of moles of each gas present in the container. Step

2: Use the ideal gas law to determine the partial pressure of each gas. Step 3:

Calculate the total pressure in the container.

Step 1: Calculate the number of moles of each gas present in the container.

1. Calculate the number of moles of nitrogen gas (N2):

moles of N2=mass

molar mass =1.00 g

28.0 g/mol = 0.0357 mol

2. Calculate the number of moles of oxygen gas (O2):

moles of O2=mass

molar mass =1.50 g

32.0 g/mol = 0.0469 mol

Step 2: Use the ideal gas law to determine the partial pressure of each gas.

The ideal gas law is given by:

P V =nRT

where: P= pressure (in atm) V= volume (in L) n= number of moles R=

universal gas constant T= temperature (in K)

1. For nitrogen gas:

PN2=n·R·T

V=0.0357 mol ·0.0821 L atm mol−1K−1·(25 + 273) K

2.00 L = 0.466 atm

2. For oxygen gas:

PO2=n·R·T

V=0.0469 mol ·0.0821 L atm mol−1K−1·(25 + 273) K

2.00 L = 0.612 atm

Step 3: Calculate the total pressure in the container. Total pressure =

PN2+PO2Total pressure = 0.466 atm + 0.612 atm Total pressure = 1.08 atm

Therefore, the partial pressure of nitrogen gas is 0.466 atm, the partial pres-

sure of oxygen gas is 0.612 atm, and the total pressure in the container is 1.08

atm.

Question 23

Question

When 10.0 grams of ethanol (C2H5OH) is burned in oxygen gas according to

the following reaction:

C2H5OH(l)+3O2(g)→2CO2(g)+3H2O(l)

how many moles of oxygen gas are required?

Solution

Step 1: Calculate the molar mass of ethanol. The molar mass of C2H5OH can

be calculated by summing the molar masses of its constituent atoms:

2×Atomic mass of C + 6 ×Atomic mass of H + 1 ×Atomic mass of O

Step 2: Convert the mass of ethanol from grams to moles. Use the molar

mass of ethanol calculated in step 1 to convert grams of ethanol to moles.

Step 3: Determine the mole ratio between ethanol and oxygen. From the

balanced chemical equation, we can see that 1 mole of ethanol requires 3 moles

of oxygen.

Step 4: Calculate the moles of oxygen required. Multiply the moles of

ethanol by the mole ratio of oxygen to ethanol.

Step 1:

The molar mass of ethanol (C2H5OH) is:

2×12.01 g/mol + 6 ×1.01 g/mol + 1 ×16.00 g/mol = 46.08 g/mol

Step 2:

Converting the mass of ethanol to moles:

Moles of C2H5OH =10.0 g

46.08 g/mol ≈0.217 moles

Step 3:

The mole ratio between ethanol and oxygen is 1:3. This means that 1 mole of

ethanol reacts with 3 moles of oxygen.

Step 4:

Calculating the moles of oxygen required:

Moles of O2= 0.217 moles ×3 moles O2

1 mole C2H5OH = 0.651 moles

Therefore, 0.651 moles of oxygen gas are required to burn 10.0 grams of

ethanol.

Question 24

Question

A mixture of 3.00 L of oxygen gas at STP and 5.00 L of hydrogen gas at 27

and 1.20 atm is allowed to react. The reaction is given by the equation:

2 H2(g) + O2(g)→2 H2O(g)

Calculate the total volume of the reaction vessel after the reaction is completed,

assuming the reaction goes to completion.

Solution

Step 1: Calculate the moles of oxygen gas present. Given:

Volume of oxygen gas (V) = 3.00 L

Pressure of oxygen gas (P) = 1.00 atm

Temperature of oxygen gas (T) = 273 K

Using the ideal gas law P V =nRT , we can calculate the moles of oxygen

gas:

n=P V

=(1.00 atm)(3.00 L)

0.0821 atm ·L/mol ·K·273 K

≈0.122 moles

Step 2: Calculate the moles of hydrogen gas present. Given:

Volume of hydrogen gas (V) = 5.00 L

Pressure of hydrogen gas (P) = 1.20 atm

Temperature of hydrogen gas (T) = 27C= 300 K

Using the ideal gas law, we can calculate the moles of hydrogen gas:

n=P V

=(1.20 atm)(5.00 L)

0.0821 atm ·L/mol ·K·300 K

≈0.243 moles

Step 3: Determine the limiting reactant. From the balanced chemical equa-

tion, we see that 1 mole of O2 reacts with 2 moles of H2. Therefore, the limiting

reactant is the one that produces the least amount of water, which in this case

is oxygen.

Step 4: Calculate the moles of water produced. Since oxygen is the limiting

reactant, all of it will react with hydrogen to produce water. The reaction

consumes 0.122 moles of oxygen, producing 0.244 moles of water.

Step 5: Calculate the volume of water vapor formed. Based on the stoichiom-

etry of the reaction, every 2 moles of water vapor occupies the same volume as 1

mole of oxygen. Thus, the total volume of water vapor formed can be calculated

as:

Volume of water vapor = 2

1×Volume of oxygen

= 2 ×3.00 L

= 6.00 L

Therefore, the total volume of the reaction vessel after the reaction is com-

pleted is 6.00 L.

Question 25

Question

A gaseous compound containing only hydrogen and nitrogen is analyzed and

found to be 85.7

(Given: molar volume at STP = 22.4 L/mol, R= 0.0821 atm ·L/mol ·K)

Solution

Step 1: Calculate the molar mass of the compound using the percent composi-

tion of nitrogen.

molar mass of N = 14.01 g/mol

molar mass of H = 1.01 g/mol

molar mass of compound = 100 g/mol = 85.7

14.01 +14.3

1.01

= 684.51 g/mol

Step 2: Use the ideal gas law to find the number of moles of the compound

present in the sample.

P V =nRT

n=P V

RT =(755 mmHg ×0.825 L)

(0.0821 atm ·L/mol ·K×298 K)

= 0.2780 mol

Step 3: Calculate the number of moles of nitrogen and hydrogen in the

compound and determine the empirical formula.

moles of N = 0.857 ×0.2780 = 0.2381 mol

moles of H = (1 −0.857) ×0.2780 = 0.0399 mol

Dividing by the smallest value gives the empirical formula NH.

Step 4: Determine the molecular formula using the molar mass calculated

in Step 1.

molar mass of empirical formula (NH) = 68 g/mol

molecular formula = 684.51

68 = 10

Therefore, the molecular formula of the compound is NH.

Question 26

Question

A 2.00 L container is filled with 5.00 mol of helium gas (He) at 25

C. If the

container is then heated to 80

C, what will be the new pressure inside the

container? (Assume ideal gas behavior)

Solution

Step 1: Calculate the initial pressure inside the container using the ideal gas

law:

P V =nRT

where: P= pressure (in atm) V= volume (in L) n= moles of gas R= ideal

gas constant = 0.0821 atm L / mol K T= temperature (in K)

Given: V= 2.00 L, n= 5.00 mol, T= 25

C = 298 K

Substitute these values into the ideal gas law:

P1=nRT1

P1=(5.00 mol)(0.0821 atm L/mol K)(298 K)

2.00 L

P1= 61.19 atm

Step 2: Calculate the final pressure inside the container using the new tem-

perature: Given: new temperature T2= 80

C = 353 K

Since the gas is kept at constant volume, we can use the following formula:

=P2

Substitute the initial pressure and temperature, and solve for the final pres-

sure P2:61.19

298 =P2

353

P2= 69.85 atm

Therefore, the new pressure inside the container when heated to 80

C will

be 69.85 atm.

Question 27

Question

When 4.50 L of ethylene gas (C2H4) reacts with excess oxygen gas in a com-

bustion reaction, 8.76 L of carbon dioxide gas (CO2) is produced. If all gas

volumes are measured at the same temperature and pressure, what is the bal-

anced chemical equation for the reaction?

Solution

Step 1: Write the balanced chemical equation for the reaction. Step 2: Use the

ideal gas law to find the number of moles of each reactant and product. Step

3: Determine the stoichiometry of the reaction based on the mole ratio of the

reactants and products. Step 4: Write the balanced chemical equation.

Step 1: Write the balanced chemical equation for the reaction.

Let the balanced chemical equation for the reaction be:

aC2H4+bO2→cCO2+dH2O

Step 2: Use the ideal gas law to find the number of moles of each reactant

and product.

Given: Volume of C2H4= 4.50 L Volume of CO2produced = 8.76 L

Since all gases are measured at the same temperature and pressure, their

volumes are directly proportional to the number of moles.

V olume1

V olume2

=Moles1

Moles2

Therefore, the number of moles of CO2produced is twice the number of

moles of C2H4consumed.

Step 3: Determine the stoichiometry of the reaction based on the mole ratio

of the reactants and products.

From the balanced chemical equation: - The coefficient in front of C2H4is

1. - The coefficient in front of CO2is c (which we don’t know yet).

This means that the coefficient in front of CO2should be 2 in order to

balance the equation.

Therefore, the balanced chemical equation is:

C2H4+ 3O2→2CO2+ 2H2O

Step 4: Write the balanced chemical equation.

The balanced chemical equation for the reaction is:

C2H4+ 3O2→2CO2+ 2H2O

Question 28

Question

A gaseous hydrocarbon is burned completely in oxygen gas to produce carbon

dioxide and water. If 0.85 moles of carbon dioxide are produced, how many

moles of the hydrocarbon were burned?

Solution

Step 1: Write the balanced chemical equation for the combustion of the hydro-

carbon.

Hydrocarbon + O2→CO2+ H2O

Step 2: Determine the molar ratio between carbon dioxide and the hydro-

carbon from the balanced chemical equation. From the equation, we see that

one mole of hydrocarbon produces one mole of carbon dioxide.

Step 3: Use the molar ratio to find the number of moles of the hydrocarbon

burned. Since 0.85 moles of carbon dioxide were produced, then 0.85 moles of

the hydrocarbon were burned as well.

Answer: 0.85 moles of the hydrocarbon were burned.

Question 29

Question

A mixture of 0.500 mol of methane (CH4) and 0.800 mol of oxygen (O2) is

placed in a container and ignited to produce carbon dioxide (CO2) and water

(H2O) according to the following unbalanced equation:

CH4+ O2→CO2+ H2O

Calculate the maximum amount of water (in grams) that can be produced from

this reaction.

Solution

Step 1: Write the balanced chemical equation for the reaction.

CH4+ 2O2→CO2+ 2H2O

Step 2: Determine the limiting reactant.

Given the mixture contains 0.500 mol of CH4and 0.800 mol of O2, we can

calculate the theoretical yield of the products for each reactant separately.

For CH4: - 0.500 mol of CH4will produce 2 * 0.500 mol = 1.000 mol of H2O

For O2: - 0.800 mol of O2will produce 2 * 0.800 mol = 1.600 mol of H2O

Since CH4produces a lower amount of water (1.000 mol), it is the limiting

reactant.

Step 3: Calculate the maximum amount of water produced.

The molar mass of water (H2O) is 18.015 g/mol.

Mass of water = 1.000 mol ×18.015 g/mol = 18.015 g

Therefore, the maximum amount of water that can be produced from this

reaction is 18.015 grams.

Question 30

Question

A gaseous hydrocarbon compound contains only carbon and hydrogen. When

6.80 g of this compound is burned in excess oxygen, 21.6 g of carbon dioxide

is produced. Determine the molecular formula of the compound assuming it is

composed of only carbon and hydrogen.

Solution

Step 1: Write the balanced chemical equation for the combustion of the hydro-

carbon compound:

Hydrocarbon + O2→CO2+ H2O

Step 2: Calculate the moles of carbon dioxide produced: Given mass of car-

bon dioxide = 21.6 g Molar mass of carbon dioxide (CO2) = 12.01 + 2(16.00) =

44.01 g/mol

Number of moles of carbon dioxide = 21.6 g

44.01 g/mol ≈0.491 mol

Step 3: Determine the moles of carbon in the hydrocarbon: 1 mol of hydro-

carbon (CxHy) produces 1 mol of CO2(from the balanced equation) So, the

number of moles of carbon in the hydrocarbon = 0.491 mol

Step 4: Calculate the mass of carbon in the hydrocarbon: Molar mass of

carbon (C) = 12.01 g/mol

Mass of carbon in 6.80 g of the hydrocarbon = 6.80 g×0.491 mol

1 mol ×12.01 g/mol ≈

39.44 g

Step 5: Determine the mass of hydrogen in the hydrocarbon: Mass of hy-

drogen in the hydrocarbon = Total mass of the hydrocarbon - Mass of carbon

Mass of hydrogen = 6.80 g - 39.44 g = 32.36 g

Step 6: Calculate the moles of hydrogen in the hydrocarbon: Molar mass of

hydrogen (H) = 1.01 g/mol

Number of moles of hydrogen = 32.36 g

1.01 g/mol ≈32.04 mol

Step 7: Find the ratio of moles of carbon to hydrogen: Divide the moles of

carbon by the moles of hydrogen: 0.491 mol

32.04 mol ≈0.0153

Step 8: Determine the empirical formula of the hydrocarbon compound: The

empirical formula of the compound is CH(1

0.0153 )≈CH65

Step 9: Calculate the molecular formula of the hydrocarbon: Given the

molecular formula mass, MM, is 21.6 g ⇒MMCH65 = 12.01 + 1(1.01) =

13.02 g/mol

The molecular formula is thus CH65 ×21.6 g/mol

13.02 g/mol ≈CH111

Question 31

Question

A reaction occurs between 5.00 L of propane gas (C3H8) and 9.00 L of oxygen

gas (O2) at a certain temperature and pressure. The reaction forms carbon

dioxide (CO2) and water vapor (H2O). If all of the propane is consumed in the

reaction, what volume of carbon dioxide is produced? Assume all gases are at

the same temperature and pressure.

Solution

Step 1: Write the balanced chemical equation for the reaction between propane

and oxygen. The balanced chemical equation for the complete combustion of

propane is:

C3H8+ 5O2−→ 3CO2+ 4H2O

Step 2: Determine the limiting reactant. To determine the limiting reactant,

we need to first calculate the number of moles of each reactant. Given: Volume

of C3H8= 5.00 L Volume of O2= 9.00 L

Using the ideal gas law, we can convert the volume of gases to moles:

n=P V

Assuming temperature and pressure are constant, we get:

nC3H8=P V

RT =(5)(V)

nO2=P V

RT =(9)(V)

Step 3: Determine the limiting reactant. The stoichiometry of the reaction

indicates that 1 mol of C3H8reacts with 5 mol of O2to produce 3 mol of CO2.

Calculate the moles of C3H8and O2:

Moles of C3H8=5

Moles of O2=9

Since the reaction requires 5 moles of O2for every mole of C3H8,O2is the

limiting reactant.

Step 4: Calculate the volume of CO2produced. From the balanced chemical

equation, 1 mole of C3H8produces 3 moles of CO2. Since O2is the limiting

reactant, we can determine the moles of CO2produced using the stoichiometry:

Moles of CO2= (3)( 9

RT ) = 27

Now, we can calculate the volume of CO2using the ideal gas law:

VCO2=27V

Question 32

Question

A 2.00 L flask contains 1.00 atm of N2, 2.00 atm of H2, and 3.00 atm of NH3

at equilibrium in the reaction:

N2(g) + 3H2(g)⇌2NH3(g)

If the temperature is held constant, will the pressure change if an additional

0.500 atm of H2is added to the flask? Justify your answer.

Solution

Step 1: Write down the balanced chemical equation for the reaction.

The balanced chemical equation is:

N2(g) + 3H2(g)⇌2NH3(g)

Step 2: Determine the initial moles and partial pressures of the gases.

Using the ideal gas law P V =nRT , we can calculate the initial moles of gas

in the flask as follows:

For N2:

n(N2) = P V

RT =(1.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

For H2:

n(H2) = P V

RT =(2.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

For NH3:

n(NH3) = P V

RT =(3.00 atm ×2.00 L)

(0.0821 atm ·L/mol ·K) ×(T)

Step 3: Check the reaction quotient Q and compare it to the equilibrium

constant K.

The reaction quotient Q is given by:

Q=[NH3]2

[N2][H2]3

If Q = K, the system is at equilibrium; if Q ¿ K, the reaction shifts left; if

Q ¡ K, the reaction shifts right.

Step 4: Determine the final moles and partial pressures of the gases.

When 0.500 atm of H2is added to the flask, the partial pressure of H2will

increase. This will cause Q to become greater than K, prompting the reaction

to shift to the left in order to re-establish equilibrium. This shift will result in

changes to the moles and partial pressures of the gases in the flask.

Question 33

Question

A reaction occurs between 12.0 L of hydrogen gas at STP and excess nitrogen

gas to produce ammonia gas. If the reaction is 85

Solution

Step 1: Write the balanced chemical equation for the reaction between hydrogen

and nitrogen to produce ammonia.

3H2(g)+N2(g)→2NH3(g)

Step 2: Determine the moles of hydrogen gas present using the ideal gas law,

P V =nRT .

PV = nRT

(1 atm)(12.0 L) = n ×(0.08206 L atm / mol K) ×(273 K)

n = (1)(12.0)

(0.08206)(273) = 0.532 mol

Step 3: Calculate the theoretical yield of ammonia gas based on the number

of moles of hydrogen gas used (using the stoichiometry of the balanced equation).

Moles of NH3=(0.85)(0.532)

3= 0.150 mol

Step 4: Convert the moles of ammonia gas to volume using the ideal gas

law.

PV = nRT

V = nRT

P=(0.150)(0.08206)(273)

1= 3.958 L

Therefore, the volume of ammonia gas that can be produced is 3.958 L.

Question 34

Question

Magnesium reacts with hydrochloric acid to produce magnesium chloride and

hydrogen gas. If 3.00 g of magnesium is reacted with excess hydrochloric acid

to produce 0.500 L of hydrogen gas at 298 K and 1.00 atm, what is the percent

yield of this reaction? (Assume the reaction goes to completion.)

Solution

Step 1: Write the balanced chemical equation for the reaction:

Mg + 2HCl →MgCl2+ H2

Step 2: Calculate the moles of magnesium used: Given: mass = 3.00 g,

molar mass of Mg = 24.31 g/mol

moles of Mg = 3.00 g

24.31 g/mol = 0.1235 mol

Step 3: Determine the moles of hydrogen gas produced using stoichiometry:

From the balanced equation, 1 mole of Mg produces 1 mole of H2 gas.

moles of H2= 0.1235 mol

Step 4: Calculate the volume of hydrogen gas produced using the ideal gas

law: Given: volume = 0.500 L, temperature = 298 K, pressure = 1.00 atm,

R=0.0821 L ·atm/K·mol

moles of H2=pressure ×volume

R×temperature =1.00 atm ×0.500 L

0.0821 L ·atm/K·mol ×298 K ≈0.0201 mol

Step 5: Calculate the theoretical yield of hydrogen gas: Theoretical yield is

the lower of the two quantities found in steps 3 and 4.

theoretical yield = 0.0201 mol

Step 6: Calculate the percent yield: Given: actual yield = 0.0201 mol,

theoretical yield = 0.0201 mol

percent yield = actual yield

theoretical yield×100% = 0.0201 mol

0.0201 mol×100% = 100%

Therefore, the percent yield of this reaction is 100

Question 35

Question

A reaction between methane gas (CH4) and oxygen gas (O2) produces carbon

dioxide gas (CO2) and water vapor (H2O) according to the following balanced

chemical equation:

CH4+ 2O2→CO2+ 2H2O

If 3.50 moles of methane gas react with excess oxygen gas, how many moles

of water vapor are produced?

Solution

Step 1: Determine the mole ratio between methane and water vapor using the

balanced chemical equation. The coefficient of CH4is 1 and the coefficient of

H2Ois 2. Therefore, the mole ratio between CH4and H2Ois 1:2.

Step 2: Calculate the number of moles of H2Oproduced. Given that 3.50

moles of CH4are reacting, we use the mole ratio to find the moles of H2O

produced:

3.50 moles CH4

1×2 moles H2O

1 mole CH4

= 7.00 moles H2O

Therefore, 7.00 moles of water vapor are produced when 3.50 moles of

methane gas react.

CHEM 131 - ADVANCED GENERAL CHEMISTRY I Gas Stoichiometry Question Bank Set 1

Students also viewed