CHEM 131 - ADVANCED GENERAL CHEMISTRY I Gas Stoichiometry Question Bank Set 4

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Gas Stoichiometry

Question Bank - Set 4

Liberty University

Question 1

Question

A reaction takes place between carbon monoxide gas (CO) and oxygen gas (O2)

to form carbon dioxide gas (CO2) according to the following balanced equation:

2CO(g) + O2(g)→2CO2(g)

If 3.50 moles of carbon monoxide gas react with excess oxygen gas, how many

moles of carbon dioxide gas will be produced?

Solution

Step 1: Calculate the moles of carbon dioxide produced per mole of CO. Given

the balanced chemical equation, 2 moles of CO is needed to produce 2 moles

of CO2. This means that for every mole of CO that reacts, 1 mole of CO2is

produced. Therefore, the moles of CO2formed per mole of CO is 1.

Step 2: Calculate the moles of CO2produced from 3.50 moles of CO. Since

1 mole of CO produces 1 mole of CO2, 3.50 moles of CO will produce 3.50 moles

of CO2.

Answer: 3.50 moles of CO2gas will be produced.

Question 2

Question

When 10.0 L of propane (C3H8) gas react with excess oxygen gas, how many

liters of carbon dioxide gas will be 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: C3H8+ 5O2→3CO2+ 4H2O

Step 2: Determine the mole ratio between propane and carbon dioxide from

the balanced chemical equation. 1 mol of propane produces 3 mol of carbon

dioxide.

Step 3: Calculate the number of moles of propane present:

Volume of propane given = 10.0 L

From the ideal gas law, we know that:

P V =nRT

n=P V

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

gas constant, and Tis the temperature. Since the temperature and pressure are

constant in this case, we can simplify the formula to:

n=V

22.4

n=10.0 L

22.4 L/mol = 0.4464 mol

Step 4: Use the mole ratio to calculate the number of moles of carbon dioxide

produced:

0.4464 mol propane ×3 mol CO2

1 mol propane = 1.3392 mol CO2

Step 5: Convert the number of moles of carbon dioxide to volume using the

ideal gas law:

n=P V

V=nRT

V=1.3392 mol ×0.0821 (atm ·L/mol ·K) ×273 K

1 atm = 30.3 L

Therefore, 30.3 liters of carbon dioxide gas will be produced.

Question 3

Question

In a chemical reaction, 3.00 L of hydrogen gas reacts with excess nitrogen gas

to produce 2.00 L of ammonia gas at the same temperature and pressure. The

balanced chemical equation for the reaction is:

3H2(g) + N2(g)→2NH3(g)

Determine the volume of hydrogen gas (in L) that would be needed to pro-

duce 10.0 L of ammonia gas under the same conditions.

Solution

Step 1: Calculate the molar ratio of hydrogen gas to ammonia gas based on the

balanced chemical equation. Step 2: Use the molar ratio to find the volume of

hydrogen gas needed to produce 10.0 L of ammonia gas.

Step 1: Calculate the molar ratio of hydrogen gas to ammonia gas.

From the balanced chemical equation, 3 moles of hydrogen gas react to

produce 2 moles of ammonia gas.

Step 2: Use the molar ratio to find the volume of hydrogen gas needed.

Given: Volume of ammonia gas produced = 10.0 L

Using the molar ratio from Step 1:

3 moles H2

2 moles NH3

=10.0 L H2

VL NH3

Solving for V:

V=10.0×2

3= 6.67 L

Therefore, 6.67 L of hydrogen gas would be needed to produce 10.0 L of

ammonia gas under the same conditions.

Question 4

Question

A reaction between hydrogen gas (H2) and nitrogen gas (N2) produces ammonia

gas (NH3) according to the equation:

3H2(g)+N2(g)→2NH3(g)

If 14.0 L of hydrogen gas reacts with an excess of nitrogen gas at STP, what

volume of ammonia gas is produced?

Solution

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

PV = nRT

Since the gases are at STP, the pressure (P) is 1 atm, the temperature (T) is

273 K, and the gas constant (R) is 0.0821 L atm/mol K.

n = PV

RT =(1 atm)(14.0 L)

(0.0821 L atm/mol K)(273 K) ≈0.787 mol

Step 2: Use the mole ratio from the balanced chemical equation to find the

moles of ammonia gas produced. From the balanced chemical equation, the

ratio of H2to NH3is 3:2.

nNH3=3

2×0.787 ≈1.18 mol

Step 3: Calculate the volume of ammonia gas produced using the ideal gas

law.

V = nRT

P=(1.18 mol)(0.0821 L atm/mol K)(273 K)

1 atm ≈27.8 L

Therefore, approximately 27.8 L of ammonia gas is produced when 14.0 L

of hydrogen gas reacts with an excess of nitrogen gas at STP.

Question 5

Question

A 2.50 L container at 25

C contains carbon monoxide gas at a pressure of 0.800

atm. If the carbon monoxide is burned to produce carbon dioxide according to

the following balanced chemical equation:

2CO(g) + O2(g)→2CO2(g)

What will be the pressure in the container after the reaction is complete

if the temperature remains constant and the volume of the container does not

change?

Solution

Step 1: Calculate the number of moles of carbon monoxide present in the con-

tainer using the ideal gas law equation:

P V =nRT

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

Ris the ideal gas constant, and Tis the temperature in Kelvin. Rearranging

the equation to solve for n:

n=P V

⇒n=(0.800 atm)(2.50 L)

(0.0821 atm ·L/mol ·K)(298 K)

⇒n≈0.082 mol

Step 2: Use the balanced chemical equation to determine the stoichiometry

between carbon monoxide and carbon dioxide. Since 2 moles of carbon monoxide

produce 2 moles of carbon dioxide, the mole ratio is 1:1.

Step 3: Calculate the number of moles of carbon dioxide produced:

moles of CO2= moles of CO

⇒moles of CO2= 0.082 mol

Step 4: Calculate the new pressure using the ideal gas law, assuming the

volume and temperature remain constant:

P2=nRT

⇒P2=(0.082 mol)(0.0821 atm ·L/mol ·K)(298 K)

2.50 L

⇒P2≈0.833 atm

Therefore, the pressure in the container after the reaction is complete will

be approximately 0.833 atm.

Question 6

Question

A mixture of ethane (C2H6) and oxygen (O2) is burned in a gas cylinder ac-

cording to the following balanced equation:

2C2H6(g) + 7O2(g)→4CO2(g) + 6H2O(g)

If 10.0 L of ethane at STP (standard temperature and pressure) and 5.00 L of

oxygen at STP are allowed to react, what volume in Liters of carbon dioxide at

STP will be produced?

Solution

Step 1: Calculate the number of moles for each reactant. Given that we have

10.0 L of ethane and 5.00 L of oxygen at STP, we can calculate the number

of moles for each using the ideal gas law, P V =nRT , where P= 1.00 atm,

V= 10.0 L (for ethane) and 5.00 L (for oxygen), T= 273 K and R= 0.0821 L ·

atm/mol ·K.

For ethane:

nC2H6=P V

RT =(1.00 atm)(10.0 L)

(0.0821 L ·atm/mol ·K)(273 K)

nC2H6=10.0

22.414 = 0.446 moles of C2H6

For oxygen:

nO2=P V

RT =(1.00 atm)(5.00 L)

(0.0821 L ·atm/mol ·K)(273 K)

nO2=5.00

22.414 = 0.223 moles of O2

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

tion, we see that the ratio of ethane to oxygen is 2:7. We can calculate the

moles of carbon dioxide produced by each reactant: Moles of CO2produced

from ethane: 0.446 moles ×4 moles CO2

2 moles C2H6= 0.892 moles CO2Moles of CO2pro-

duced from oxygen: 0.223 moles ×4 moles CO2

7 moles O2= 0.128 moles CO2

Since oxygen produces fewer moles of carbon dioxide, oxygen is the limiting

reactant.

Step 3: Calculate the volume of carbon dioxide produced. Since the reaction

ratio shows that 4 moles of CO2 are produced for every 7 moles of O2, we can

calculate the volume of CO2 produced using the ideal gas law:

VCO2=nRT

P=(0.128)(0.0821)(273)

1.00 = 2.66 L

Therefore, 2.66 Liters of carbon dioxide at STP will be produced.

Question 7

Question

A reaction takes place between 3.00 L of oxygen gas at STP and excess hydrogen

gas to produce water vapor. If the reaction yields 4.50 L of water vapor at STP,

what is the balanced chemical equation for the reaction?

Solution

Step 1: Determine the molar ratio between oxygen and water vapor based on

the given volumes at STP.

Given: - Volume of oxygen gas = 3.00 L - Volume of water vapor produced

= 4.50 L - Both volumes are at STP

From the ideal gas law, we know that 1 mole of any ideal gas at STP occupies

22.4 L. Therefore: - 3.00 L of oxygen gas at STP = 3.00 L / 22.4 L/mol = 0.134

mols of O2 - 4.50 L of water vapor at STP = 4.50 L / 22.4 L/mol = 0.201 mols

of H2O

Step 2: Determine the molar ratios of oxygen to water vapor in the balanced

chemical equation.

The reaction between oxygen and hydrogen to form water can be represented

as:

2H2+ O2→2H2O

From the balanced chemical equation, we see that 1 mol of O2 yields 2 mol

of H2O. Therefore, the molar ratio of O2 to H2O is 1:2.

Step 3: Write the balanced chemical equation for the reaction.

The balanced chemical equation is:

2H2+ O2→2H2O

Question 8

Question

A gaseous compound containing only carbon and hydrogen is combusted in

excess oxygen gas. If the combustion of 2.00 g of the compound produces 7.52

g of carbon dioxide and 2.59 g of water vapor, determine the empirical formula

of the compound.

Solution

Step 1: Find the number of moles of carbon dioxide produced. Given: mCO2=

7.52 g The molar mass of carbon dioxide (CO2) is 44.01 g/mol.

Number of moles of CO2=mCO2

Molar mass of CO2

=7.52 g

44.01 g/mol ≈0.171 mol

Step 2: Find the number of moles of water vapor produced. Given: mH2O=

2.59 g The molar mass of water (H2O) is 18.02 g/mol.

Number of moles of H2O = mH2O

Molar mass of H2O=2.59 g

18.02 g/mol ≈0.144 mol

Step 3: Determine the number of moles of carbon and hydrogen that combine

to form the compound. From the balanced combustion reaction equation, one

mole of the compound reacts to produce one mole of CO and one mole of

HO. Therefore, the compound produced 0.171 moles of carbon for every mole

of compound. Similarly, the compound produced 0.144 moles of hydrogen for

every mole of compound.

Step 4: Determine the empirical formula of the compound. Since the com-

pound contains only carbon and hydrogen, the empirical formula is CxHy. From

step 3, the compound has a ratio of 0.171 moles of carbon to 0.144 moles of hy-

drogen. To find the simplest whole number ratio, divide both values by the

smaller number of moles (0.144):

0.171 mol/0.144 mol

0.144 mol/0.144 mol =1.188

1≈6

This means the compound has an empirical formula of C6H5.

Question 9

Question

A gaseous compound made up of nitrogen and chlorine has a molar volume of

50.8 L/mol at STP (0

C and 1 atm). If the compound contains 25

Solution

Step 1: Determine the masses of nitrogen and chlorine in 1 mole of the com-

pound. Let’s assume we have 100 g of the compound. - Mass of nitrogen in the

100 g compound: 25- Mass of chlorine in the 100 g compound: 100 g - 25 g =

75 g

Step 2: Convert the masses of nitrogen and chlorine to moles. - Moles of

nitrogen: 25g

14.01g/mol ≈1.78 mol (using atomic mass of nitrogen = 14.01 g/mol) -

Moles of chlorine: 75g

35.45g/mol ≈2.11 mol (using atomic mass of chlorine = 35.45

g/mol)

Step 3: Determine the simplest whole number ratio of moles of nitrogen to

moles of chlorine. By dividing the moles of each element by the smaller number

of moles: - For nitrogen: 1.78mol

1.78mol ≈1 - For chlorine: 2.11mol

1.78mol ≈1.19

Step 4: Multiply the ratio by a whole number to get whole numbers for the

subscripts in the empirical formula. Since the ratio for chlorine is close to 1.19,

we can round it off: - For nitrogen: 1 ×1 = 1 - For chlorine: 1.19 ×1≈1.19 ⇒

round to 1

Therefore, the empirical formula of the compound is NCl.

Question 10

Question

A reaction between 25.0 g of methane (CH4) and 30.0 g of oxygen gas (O2)

produces carbon dioxide (CO2) and water (H2O) according to the following

balanced chemical equation:

CH4(g) + 2O2(g)→CO2(g) + 2H2O(g)

What is the limiting reactant in this reaction? How many grams of water

are produced?

Solution

Step 1: Calculate the number of moles of each reactant.



Moles of methane (CH4):

Moles = Mass

Molar Mass =25.0 g

16.05 g/mol ≈1.556 mol



Moles of oxygen gas (O2):

Moles = Mass

Molar Mass =30.0 g

32.00 g/mol ≈0.938 mol

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

we compare the moles of each reactant to the stoichiometry of the balanced

equation.



For methane: 1.556 mol CH4

1= 1.556 mol O2need for complete reaction.



For oxygen gas: 0.938 mol O2

2= 0.469 mol CH4need for complete reaction.

Since we need 1.556 mol of O2but only have 0.938 mol, oxygen gas is the

limiting reactant.

Step 3: Calculate the mass of water produced using the limiting reactant.



Moles of water produced:

Moles = Moles of limiting reactant×Coefficient of water

Coefficient of limiting reactant = 0.938 mol×2

2= 0.938 mol



Mass of water produced:

Mass = Moles ×Molar Mass = 0.938 mol ×18.015 g/mol = 16.93 g

Therefore, the limiting reactant is oxygen gas and 16.93 g of water are pro-

duced.

Question 11

Question

A sample of a gaseous compound containing only carbon and hydrogen was

burned in excess oxygen. The volume of carbon dioxide produced was found to

be 2.50 L at 25

C and 1 atm. If the initial mass of the compound was 3.00 g,

what is the empirical formula of the compound?

(Take the molar volume of a gas at STP to be 22.4 L/mol, and the molar

mass of carbon dioxide to be 44.01 g/mol.)

Solution

Step 1: Calculate the number of moles of carbon dioxide produced. The volume

of carbon dioxide produced is 2.50 L. From the ideal gas law, we can calculate

the number of moles of carbon dioxide using the formula:

n=P V

Substitute P= 1 atm, V= 2.50 L, R= 0.0821 L·atm/mol·K, and T=

25C= 298 K into the formula:

n=(1 atm)(2.50 L)

(0.0821 L ·atm/mol ·K)(298 K)

n≈0.103 mol

Step 2: Calculate the number of moles of the compound. From the balanced

chemical equation, we know that the compound reacts to produce 1 mol of car-

bon dioxide. Since 1 mol of carbon dioxide is produced from 1 mol of compound,

the number of moles of the compound is also 0.103 mol.

Step 3: Calculate the molar mass of the compound. Given that the initial

mass of the compound is 3.00 g, and the number of moles of the compound is

0.103 mol, we can calculate the molar mass using the formula:

Molar mass = Total mass

Number of moles

Molar mass = 3.00 g

0.103 mol

Molar mass ≈29.13 g/mol

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

formula of the compound can be found by dividing the molar mass by the molar

mass of each element and then multiplying by the smallest whole number ratio.

To do this, we need to determine the number of moles of carbon and hydrogen

in the compound. - Carbon: Molar mass of carbon is about 12.01 g/mol, so the

number of moles of carbon in the compound is:

Moles of carbon = 12.01

29.13 ×0.103

- Hydrogen: Since the compound only contains carbon and hydrogen, the

rest of the molar mass is due to hydrogen. Therefore, the number of moles of

hydrogen in the compound is:

Moles of hydrogen = 1−12.01

29.13×0.103

Step 5: Determine the empirical formula. From the moles of carbon and

hydrogen, we can determine the empirical formula by dividing each mole value

by the smallest mole value obtained. This calculation should give us a ratio of

whole numbers. Thus, the empirical formula would be CaHbwith appropriate

values for a and b.

Question 12

Question

A gaseous compound containing only carbon and hydrogen is burned in excess

oxygen gas producing 5.0 L of carbon dioxide gas at 1.0 atm and 25

C. If the

molar volume of a gas at STP is 22.4 L/mol, determine the empirical formula

of the compound.

Solution

Step 1: Write the balanced chemical equation for the combustion of the gaseous

compound:

CmHn+ (m+n/4)O2→mCO2+n/2H2O

Step 2: Calculate the moles of carbon dioxide produced using the ideal gas

law: Given: V= 5.0 L, P= 1.0 atm, T= 25C= 298 K

nCO2=P V

RT =(1.0 atm)(5.0 L)

0.0821 atm ·L/mol ·K·(298 K)

nCO2≈0.203 moles

Step 3: Determine the moles of carbon in the compound by comparing with

moles of carbon dioxide produced: From the balanced equation, 1 mole of CmHn

produces mmoles of CO2So, moles of carbon in CmHn=m×nCO2

Step 4: Determine the moles of hydrogen in the compound by comparing

with moles of water produced: From the balanced equation, 1 mole of CmHn

produces n/2 moles of H2OSo, moles of hydrogen in CmHn= (n/2) ×nCO2

Step 5: Determine the simplest whole number ratio of carbon and hydrogen:

Divide the moles of carbon and hydrogen by the smallest of the two values

obtained in Steps 3 and 4. This will give you the empirical formula.

Step 6: Write the empirical formula for the compound: The empirical for-

mula would be CxHywhere x and y are the whole numbers obtained from the

ratio in Step 5.

Question 13

Question

When 2.0 L of hydrogen gas reacts with excess oxygen gas, water vapor is

produced. How many liters of water vapor at 25

C and 1 atm are produced if

the reaction goes to completion?

Solution

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

balanced chemical equation for the reaction of hydrogen gas with oxygen gas to

produce water vapor is:

2H2(g) + O2(g)→2H2O(g)

Step 2: Determine the stoichiometry of the reaction. From the balanced

chemical equation, we can see that 2 moles of hydrogen gas react with 1 mole

of oxygen gas to produce 2 moles of water vapor.

Step 3: Use the ideal gas law to calculate the volume of water vapor pro-

duced. Given: - Volume of hydrogen gas (V): 2.0 L - Temperature (T): 25

C =

298 K - Pressure (P): 1 atm

Using the ideal gas law:

P V =nRT

where: P = pressure in atm V = volume in liters n = number of moles R =

ideal gas constant = 0.0821 L

atm/(K

mol) T = temperature in Kelvin

Step 4: Calculate the number of moles of hydrogen gas. Since the volume

of hydrogen gas is given as 2.0 L, we can use the ideal gas law to calculate the

number of moles of hydrogen gas.

n=P V

RT =(1.0atm)(2.0L)

(0.0821L·atm/(K·mol)·298K)

Step 5: Calculate the number of moles of water vapor produced. From the

stoichiometry of the reaction, we know that 2 moles of water vapor are produced

for every 2 moles of hydrogen gas. Thus, the number of moles of water vapor

produced is equal to the number of moles of hydrogen gas.

Step 6: Use the ideal gas law to calculate the volume of water vapor pro-

duced. Using the ideal gas law:

n=P V

Substitute the values:

V=nRT

P=(2.0mol)(0.0821L·atm/(K·mol)·298K)

1.0atm

V= 48.92L

Therefore, 48.92 liters of water vapor are produced at 25

C and 1 atm when

2.0 L of hydrogen gas reacts with excess oxygen gas.

Question 14

Question

A mixture of carbon monoxide and hydrogen gases reacts to produce methanol

according to the following balanced equation:

CO(g)+2H2(g)→CH3OH(g)

If 3.00 L of CO gas at 1.25 atm and 300 K is reacted with an excess of

hydrogen gas, how many grams of methanol can be produced in the reaction?

Solution

Step 1: Calculate the moles of CO. Given: Volume of CO gas, VCO = 3.00 L

Pressure of CO gas, PCO = 1.25 atm Temperature, T= 300 K

Using the ideal gas law:

nCO =PCOVCO

where R= 0.0821 L ·atm/K·mol.

nCO =(1.25 atm)(3.00 L)

0.0821 L ·atm/K·mol ·300 K

nCO =3.75

24.63

nCO ≈0.152 mol

Step 2: Use stoichiometry to find the moles of methanol produced. From

the balanced chemical equation, it is seen that 1 mole of CO produces 1 mole of

methanol. Therefore, 0.152 moles of CO will produce 0.152 moles of methanol.

Step 3: Convert moles of methanol to grams. Given: Molar mass of methanol,

MCH3OH = 32.04 g/mol

Mass of CH3OH = nCH3OH ×MCH3OH = 0.152 mol ×32.04 g/mol

Mass of CH3OH = 4.87 g

Therefore, approximately 4.87 grams of methanol can be produced in the

reaction.

Question 15

Question

A mixture of acetylene gas (C2H2) and oxygen gas (O2) is used for welding. If

3.00 L of acetylene gas at STP (273 K and 1 atm) is mixed with 5.00 L of oxygen

gas at 87

C and 0.850 atm, determine the limiting reactant and calculate the

volume of carbon dioxide gas (CO2) produced at STP.

Solution

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

and oxygen:

2C2H2+ 5O2→4CO2+ 2H2O

Step 2: Determine the number of moles of acetylene gas using the ideal gas

law:

n=P V

nC2H2=(1 atm)(3.00 L)

0.0821 L ·atm/mol ·K·(273 K)

nC2H2≈0.111 mol

Step 3: Determine the number of moles of oxygen gas using the ideal gas

law:

nO2=(0.850 atm)(5.00 L)

0.0821 L ·atm/mol ·K·(360 K)

nO2≈0.180 mol

Step 4: Calculate the mole ratio for the reaction: From the balanced equa-

tion, the mole ratio of C2H2to O2is 2:5.

Step 5: Determine the limiting reactant: Since the mole ratio of C2H2to

O2is 2:5, the limiting reactant is acetylene (C2H2) since there are fewer moles

available.

Step 6: Calculate the volume of carbon dioxide gas produced at STP: Using

the mole ratio from the balanced equation: 1 mol of C2H2produces 4 mol of

CO2.

Moles of CO2= 0.111 mol ×4

2= 0.222 mol

Now, use the ideal gas law to find the volume of CO2at STP:

VCO2=nRT

P=(0.222 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm

VCO2≈4.78 L

Therefore, the volume of carbon dioxide gas produced at STP is approxi-

mately 4.78 L.

Question 16

Question

A gaseous compound contains only carbon, hydrogen, and sulfur. When a 0.250

g sample of the compound is burned in excess oxygen, 0.592 g of carbon dioxide

and 0.306 g of water are produced. Another sample of the compound, with a

mass of 0.500 g, is found to contain 0.0948 g of sulfur. Determine the empirical

formula of the compound.

Solution

Step 1: Find the moles of carbon and hydrogen in the compound



Moles of carbon dioxide produced:

moles of CO2=0.592 g

44.01 g/mol = 0.0135 mol



Moles of carbon in the compound:

moles of C= moles of CO2= 0.0135 mol



Moles of water produced:

moles of H2O=0.306 g

18.02 g/mol = 0.0169 mol



Moles of hydrogen in the compound:

moles of H= 2 ×moles of H2O= 2 ×0.0169 mol = 0.0338 mol

Step 2: Find the moles of sulfur in the compound



Moles of sulfur in the compound:

moles of S=0.0948 g

32.06 g/mol = 0.00296 mol

Step 3: Determine the empirical formula of the compound



Multiply the moles of each element by a common factor to obtain whole

number ratios.

Carbon : 0.0135 mol ×1

0.00296 mol ≈4.56 ≈5

Hydrogen : 0.0338 mol ×1

0.00296 mol ≈11.42 ≈11

Sulfur : 0.00296 mol ×1

0.00296 mol = 1



The empirical formula is thus C5H11S.

Therefore, the empirical formula of the compound is C5H11S.

Question 17

Question

A 2.00 L container holds 3.00 moles of hydrogen gas and 5.00 moles of oxygen

gas at a certain temperature and pressure. If the following reaction takes place:

2H2(g) + O2(g)→2H2O(g)

What is the limiting reactant, and how many moles of water can be formed?

Solution

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

is:

2H2(g) + O2(g)→2H2O(g)

Step 2: Determine the moles of each reactant. Given: - Moles of hydrogen

gas, n(H2)=3.00 moles - Moles of oxygen gas, n(O2)=5.00 moles

Step 3: Calculate the moles of water that can be formed from each reactant.

- From hydrogen gas: n(H2)

2=3.00

2= 1.50 moles of water can be formed. - From

oxygen gas: n(O2)

1= 5.00 moles of water can be formed.

Step 4: Identify the limiting reactant. Since hydrogen gas forms the least

amount of water (1.50 moles) compared to oxygen gas (5.00 moles), hydrogen

gas is the limiting reactant.

Step 5: Calculate the maximum moles of water that can be formed using

the limiting reactant. Since 2 moles of hydrogen gas produce 2 moles of water,

the number of moles produced by 3.00 moles of hydrogen gas is:

Number of moles of water = 3.00 moles ×2 moles of water

2 moles of hydrogen gas = 3.00 moles

Therefore, 3.00 moles of water can be formed using hydrogen gas as the

limiting reactant.

Question 18

Question

A reaction between solid aluminum and solid iodine forms solid aluminum iodide

according to the following balanced chemical equation:

2Al + 3I2→2AlI3

If 10.0 g of aluminum reacts with excess iodine, how many grams of alu-

minum iodide will be produced?

Solution

Step 1: Calculate the number of moles of aluminum using its molar mass. Given:

Mass of aluminum = 10.0 g Molar mass of aluminum = 26.98 g/mol

Number of moles of Al = 10.0 g

26.98 g/mol

Number of moles of Al ≈0.370 mol

Step 2: Use stoichiometry to find the number of moles of aluminum iodide

produced. From the balanced chemical equation, we see that 2 moles of alu-

minum produce 2 moles of aluminum iodide.

Number of moles of AlI3= Number of moles of Al ×2 mol AlI3

2 mol Al

Number of moles of AlI3= 0.370 mol ×1

Number of moles of AlI3= 0.370 mol

Step 3: Calculate the mass of aluminum iodide produced using its molar

mass. Molar mass of aluminum iodide = 2(26.98 g/mol) + 3(126.90 g/mol) =

266.82 g/mol

Mass of AlI3= Number of moles of AlI3×Molar mass of AlI3

Mass of AlI3= 0.370 mol ×266.82 g/mol

Mass of AlI3≈98.73 g

Therefore, approximately 98.73 grams of aluminum iodide will be produced.

Question 19

Question

A 4.00 L container of hydrogen gas at 2.00 atm and 25

C is reacted with excess

oxygen to produce water. If the reaction goes to completion at a constant

pressure of 2.00 atm and 25

C, what volume of water vapor is produced?

Solution

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

chemical equation for the reaction of hydrogen gas with oxygen to produce water

is:

2H2(g)+O2(g)→2H2O(g)

Step 2: Determine the moles of hydrogen gas present. Using the ideal gas

law equation P V =nRT where Pis pressure, Vis volume, nis the number

of moles, Ris the ideal gas constant, and Tis temperature in Kelvin, we can

rearrange the equation to solve for moles:

n=P V

Given: P= 2.00 atm, V= 4.00 L, T= 25◦C = 25 + 273 = 298 K, and

R= 0.0821 L ·atm/mol ·K.

Calculating:

n=(2.00 atm)(4.00 L)

(0.0821 L ·atm/mol ·K)(298 K) ≈0.33 mol

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

chemical equation, we see that 2 moles of hydrogen gas produces 2 moles of

water vapor. Therefore, 0.33 moles of hydrogen gas will produce 0.33 moles of

water vapor.

Step 4: Calculate the volume of water vapor produced. Using the ideal gas

law with the new volume, moles, pressure, and temperature:

V=nRT

Given: n= 0.33 mol, P= 2.00 atm, T= 25◦C = 298 K, and R= 0.0821 L ·

atm/mol ·K.

Calculating:

V=(0.33 mol)(0.0821 L ·atm/mol ·K)(298 K)

2.00 atm ≈4.90 L

Therefore, approximately 4.90 L of water vapor is produced.

Question 20

Question

A sample of manganese dioxide (MnO2) is added to excess hydrochloric acid

(HCl), resulting in the production of chlorine gas (Cl2). If 2.50 g of MnO2

produces 0.560 g of Cl2, what is the percent yield of the reaction?

(Note: The balanced chemical equation for the reaction is given by:

MnO2+ 4HCl →MnCl2+ Cl2+ 2H2O

)

Solution

Step 1: Calculate the molar mass of MnO2and Cl2. The molar mass of MnO2

is:

Molar mass of Mn+2×Molar mass of O = 54.94 g/mol+2×16.00 g/mol = 86.94 g/mol

The molar mass of Cl2is:

2×Molar mass of Cl = 2 ×35.45 g/mol = 70.90 g/mol

Step 2: Calculate the moles of MnO2and Cl2produced. The moles of MnO2is

calculated as:

Moles of MnO2=2.50 g

86.94 g/mol = 0.0288 mol

Using the balanced chemical equation, we see that 1 mole of MnO2produces

1 mole of Cl2. Hence, the moles of Cl2produced is also 0.0288 mol. Step 3:

Calculate the theoretical yield of Cl2. The theoretical yield can be calculated

from the moles of Cl2:

Theoretical yield of Cl2= Moles of Cl2×Molar mass of Cl2= 0.0288 mol×70.90 g/mol = 2.04 g

Step 4: Calculate the percent yield. The percent yield is calculated using the

formula:

Percent yield = Actual yield

Theoretical yield×100%

Given that the actual yield of Cl2is 0.560 g, we substitute the values into the

formula:

Percent yield = 0.560 g

2.04 g ×100% = 27.5%

Therefore, the percent yield of the reaction is 27.5

Question 21

Question

A mixture of hydrogen gas (H2) and nitrogen gas (N2) is reacted to form am-

monia gas (NH3) according to the following balanced equation:

N2(g) + 3H2(g)→2NH3(g)

If 8.00 grams of hydrogen gas and 28.0 grams of nitrogen gas are available,

what is the limiting reactant? What is the maximum mass of ammonia that

could be produced from this mixture?

(Atomic masses: H = 1.01 g/mol, N = 14.01 g/mol)

Solution

Step 1: Calculate the number of moles for each reactant. Given mass of hydrogen

gas, mH2= 8.00 g and molar mass of hydrogen, MH2= 1.01 g/mol:

Moles of H2=mH2

MH2

=8.00 g

1.01 g/mol = 7.92 mol

Given mass of nitrogen gas, mN2= 28.0 g and molar mass of nitrogen,

MN2= 14.01 g/mol:

Moles of N2=mN2

MN2

=28.0 g

14.01 g/mol = 2.00 mol

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

tion, the stoichiometry of the reaction indicates that 1 mole of N2reacts with 3

moles of H2to produce 2 moles of NH3. Let’s calculate the moles of NH3that

can be produced from the given amounts of reactants:

For hydrogen gas:

Moles of NH3(from H2) = 7.92 mol

3×2=5.28 mol

For nitrogen gas:

Moles of NH3(from N2) = 2.00 mol

1×2=4.00 mol

Since the moles of NH3from nitrogen gas is fewer, nitrogen gas is the limiting

reactant.

Step 3: Calculate the maximum mass of ammonia that could be produced.

From the balanced equation, 1 mole of NH3has a molar mass of 17.03 g.

Mass of NH3= Moles of NH3×Molar mass of NH3= 4.00 mol×17.03 g/mol = 68.12 g

Therefore, the maximum mass of ammonia that could be produced from the

mixture of 8.00 grams of hydrogen gas and 28.0 grams of nitrogen gas is 68.12

grams.

Question 22

Question

A mixture of ethylene (C2H4) and hydrogen gas (H2) is passed over a catalyst to

produce ethane (C2H6) according to the following balanced chemical equation:

C2H4(g) + H2(g)→C2H6(g)

If 10.0 L of C2H4is reacted with excess H2at STP (standard temperature

and pressure), what volume of C2H6is produced (also at STP)?

Solution

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

balanced chemical equation is:

C2H4(g) + H2(g)→C2H6(g)

Step 2: Determine the stoichiometry of the reaction. From the balanced

chemical equation, we see that the stoichiometry of the reaction is 1:1 for C2H4

to C2H6.

Step 3: Find the molar volume of gases at STP. At STP (standard temper-

ature and pressure), 1 mole of an ideal gas occupies 22.4 L.

Step 4: Calculate the number of moles of C2H4. Given: Volume of C2H4=

10.0 L At STP, 10.0 L of C2H4is equal to 10.0/22.4 = 0.4464 moles of C2H4.

Step 5: Determine the volume of C2H6produced. Since the stoichiometry

of the reaction is 1:1 for C2H4to C2H6, the volume of C2H6produced will also

be 10.0 L.

Question 23

Question

When 2.00 L of methane gas (CH4) reacts with 4.00 L of oxygen gas (O2) at

STP, the following reaction takes place:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Calculate the volume of carbon dioxide gas produced assuming the reaction

goes to completion.

Solution

Step 1: Write and balance the chemical equation for the reaction.

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

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

we need to compare the stoichiometry of methane to oxygen. 1 mol of CH4

produces 1 mol of CO2. 1 mol of O2produces 1/2 mol of CO2.

Since 1 mol of CH4produces more CO2than 1 mol of O2, methane is the

limiting reactant.

Step 3: Calculate the volume of CO2produced using the ideal gas law. From

the balanced chemical equation, we see that 1 mol of CH4produces 1 mol of

CO2. Thus, 2.00 L of CH4at STP will produce 2.00 L of CO2.

Therefore, the volume of CO2produced is 2.00 L.

Question 24

Question

Calculate the volume of carbon dioxide gas produced at STP when 15.0 grams

of calcium carbonate (CaCO3) react completely with excess hydrochloric acid

according to the following balanced chemical equation:

CaCO3+ 2HCl →CaCl2+ CO2+ H2O

Solution

Step 1: Calculate the number of moles of CaCO3. Given mass of CaCO3: 15.0

g Molar mass of CaCO3: 40.08 + 12.01 + 3(16.00) = 100.09 g/mol

Number of moles of CaCO3=15.0 g

100.09 g/mol = 0.1499 mol

Step 2: Determine the limiting reactant. Using the stoichiometry of the

balanced equation, we can see that 1 mole of CaCO3produces 1 mole of CO2.

Since the molar ratio between CaCO3and CO2is 1:1, the number of moles

of CO2produced will also be 0.1499 mol.

Step 3: Calculate the volume of CO2gas at STP. Given conditions: standard

temperature and pressure (STP) correspond to 0◦C and 1 atm.

Using the ideal gas law: P V =nRT where R= 0.0821 L ·atm/mol ·K is

the gas constant.

V=nRT

P=(0.1499 mol)(0.0821 L·atm/mol·K)(273 K)

1 atm = 3.95 L

Therefore, the volume of CO2gas produced at STP when 15.0 grams of

CaCO3react completely with excess hydrochloric acid is 3.95 L.

Question 25

Question

A 2.00 L container at 27◦C and 1.50 atm contains a mixture of methane (CH4)

and oxygen (O2). The gases react to form carbon dioxide (CO2) and water

vapor (H2O). If the final pressure in the container is 3.00 atm at 27◦C, what is

the composition (in moles) of the initial mixture?

Solution

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

CH4(g) + 2O2(g)→CO2(g) + 2H2O(g)

Step 2: Calculate the moles of gases present initially using the ideal gas law.

Given: - Initial volume V1= 2.00L- Initial pressure P1= 1.50atm - Initial

temperature T1= 27◦C= 300K- Final pressure P2= 3.00atm

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

moles of gas in the container:

n1=P1V1

RT1

n1=(1.50 atm)(2.00 L)

0.0821 L ·atm/mol ·K(300 K)

n1≈0.1225 mol

Step 3: Determine the mole ratios of the reactants and products from the

balanced chemical equation. 1 mol of CH4reacts with 2 mol of O2to produce

1 mol of CO2and 2 mol of H2O.

Step 4: From the mole ratio, calculate the moles of each gas present initially.

Let the moles of CH4be xand the moles of O2be y.

x+ 2y=n1

x= 0.1225 −2y

Step 5: Use stoichiometry to relate the moles of the reactants and products.

Since 1 molecule of CH4reacts with 2 molecules of O2, we have:

x= 2y

Step 6: Solve the system of equations from Step 4 and Step 5 to find the

moles of each gas. Substitute x= 2yinto the equation x= 0.1225 −2y:

2y= 0.1225 −2y

4y= 0.1225

y= 0.030625 mol

Substitute y= 0.030625 back into x= 2y:

x= 2(0.030625)

x= 0.06125 mol

Therefore, the initial composition of the mixture is 0.06125 mol of CH4and

0.030625 mol of O2.

Question 26

Question

A 2.00 L container at 25

C and 1.00 atm pressure contains a mixture of equal

masses of neon gas and argon gas. If all of the neon is burned to form neon

fluoride gas according to the equation:

2Ne(g) + 5F2(g)→2NeF5(g)

what is the total pressure in the container after the reaction is complete?

Given:



Molar masses: Ne = 20.18 g/mol, Ar = 39.95 g/mol, F = 19.00 g/mol



1 atm = 101.3 kPa

Solution

Step 1: Calculate the number of moles of neon and argon in the container before

the reaction.

The molar mass of neon (Ne) is 20.18 g/mol, so the number of moles of neon

is:

moles of Ne = mass of Ne

molar mass of Ne =1

20.18 = 0.0495 mol

Since the mass of argon (Ar) is the same as neon, the number of moles of argon

is also 0.0495 mol.

Step 2: Determine which reactant is limiting and the maximum moles of

NeF5 that can be produced.

From the balanced chemical equation, we see that 2 moles of Ne react with

5 moles of F2 to produce 2 moles of NeF5. This means that 1 mole of Ne will

require 5/2 = 2.5 moles of F2 to react completely.

Since we have an equal number of moles of neon (0.0495 mol) and argon

(0.0495 mol) but only half a mole of fluorine, the limiting reactant is fluorine.

Step 3: Calculate the moles of NeF5 produced.

Using the ratio from the balanced chemical equation:

moles of NeF5= 0.0495 mol Ne ×2

2= 0.0495 mol NeF5

Step 4: Calculate the total pressure in the container after the reaction.

To find the total moles of gas in the container after the reaction, we sum the

moles of neon, argon, and NeF5:

Total moles of gas = 0.0495 mol Ne+0.0495 mol Ar+0.0495 mol NeF5= 0.1485 mol

Using 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:

P=nRT

V=(0.1485)(0.0821)(298)

2.00 = 1.22 atm

Converting this pressure to kilopascals (kPa):

1.22 atm ×101.3 kPa/atm = 123.8 kPa

Question 27

Question

A gaseous compound is formed by the reaction of 4.0 L of hydrogen gas with

3.0 L of nitrogen gas. The balanced chemical equation is given by:

3H2(g) + N2(g)→2NH3(g)

If the reaction goes to completion at a constant temperature and pressure,

what volume of ammonia gas is produced?

Solution

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

The balanced chemical equation is:

3H2(g) + N2(g)→2NH3(g)

Step 2: Determine the mole ratios from the balanced chemical equation.

From the balanced chemical equation, we can see that 3 moles of hydrogen

react with 1 mole of nitrogen to produce 2 moles of ammonia.

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

reaction.

Given that 4.0 L of hydrogen is used in the reaction, we can use the ideal

gas law to find the number of moles:

P V =nRT

n=P V

RT =(1.0atm)(4.0L)

(0.0821 L·atm/mol ·K)(298 K)≈0.16 mol

Similarly, for nitrogen:

n=(1.0atm)(3.0L)

(0.0821 L·atm/mol ·K)(298 K)≈0.12 mol

Step 4: Determine the limiting reactant.

To find the limiting reactant, we need to compare the number of moles

of each reactant and their stoichiometric coefficients in the balanced chemical

equation. It turns out that the limiting reactant is nitrogen gas.

Step 5: Calculate the number of moles of ammonia gas produced using the

limiting reactant.

Since nitrogen is the limiting reactant, it will be completely consumed in

the reaction. Using the mole ratio from the balanced chemical equation, we can

calculate the number of moles of ammonia produced:

moles of NH3= (0.12 mol N2)×2mol NH3

1mol N2

= 0.24 mol NH3

Step 6: Calculate the volume of ammonia gas produced.

Finally, we can use the ideal gas law to find the volume of ammonia gas

produced:

P V =nRT

V=nRT

P=(0.24 mol)(0.0821 L·atm/mol ·K)(298 K)

1.0atm ≈5.8L

Therefore, approximately 5.8 L of ammonia gas is produced during the re-

action.

Question 28

Question

A gaseous compound containing only carbon and hydrogen was analyzed and

found to be 85.6

Solution

Step 1: Determine the molar mass of carbon and hydrogen. Let us assume we

have 100 g of the compound.

Given that the compound is 85.6- Mass of carbon in 100 g of the compound

= 85.6 g - Mass of hydrogen in 100 g of the compound = 14.4 g

The molar mass of carbon is approximately 12 g/mol and the molar mass of

hydrogen is approximately 1 g/mol.

Step 2: Calculate the number of moles of each element present. - Number of

moles of carbon = 85.6 g / 12 g/mol 7.133 mol - Number of moles of hydrogen

= 14.4 g / 1 g/mol = 14.4 mol

Step 3: Determine the empirical formula of the compound. To find the

empirical formula, we need to find the simplest whole-number ratio of carbon to

hydrogen. Divide the number of moles of each element by the smallest number

of moles (7.133 mol in this case) to obtain the ratio of atoms.

Dividing by 7.133 gives: - Carbon: 7.133 mol / 7.133 1 - Hydrogen: 14.4

mol / 7.133 2

This indicates the empirical formula of the compound is CH2.

Step 4: Calculate the empirical formula mass. The empirical formula mass

of CH2 = 12 g/mol (C) + 2(1 g/mol) (H) = 14 g/mol

Step 5: Determine the molecular formula of the compound. Given that the

molar mass of the compound is 58.12 g/mol, calculate the factor by which the

empirical formula mass must be multiplied to obtain the molar mass.

Factor = 58.12 g/mol

14 g/mol ≈4.150

Therefore, the molecular formula of the compound is:

1×C= C

2×H= H2

So, the molecular formula of the compound is C4H8.

Question 29

Question

A gaseous compound containing carbon and hydrogen is burned in excess oxygen

gas. If 5.00 L of the compound at STP (standard temperature and pressure)

produces 11.2 g of CO2and 4.60 g of H2O, what is the empirical formula of the

compound?

Solution

Step 1: Find the number of moles of CO2and H2O produced. Given: - Volume

of gaseous compound = 5.00 L - Mass of CO2produced = 11.2 g - Mass of H2O

produced = 4.60 g

First, find the number of moles of CO2and H2O: - Moles of CO2=

11.2 g

44.01 g/mol = 0.2547 mol - Moles of H2O = 4.60 g

18.015 g/mol = 0.2555 mol

Step 2: Determine the limiting reactant. To find the limiting reactant, we

need to compare the amount of CO2and H2O produced with the stoichiometry

of the balanced reaction. The balanced reaction for complete combustion of a

hydrocarbon looks like this:

CaHb+ (a + b/4)O2→aCO2+ b/2H2O

From the equation, we see that each mole of hydrocarbon produces 1 mole of

CO2and b/2 moles of H2O.

Now, let’s calculate the theoretical moles of CO2and H2O for the given

moles of compound: - For CO2:0.2547 mol

1= 0.2547 mol - For H2O: 0.2555 mol

0.2555 mol

2= 0.12775 mol

Since the moles of CO2produced exceed the theoretical moles, CO2is the

excess reactant and H2O is the limiting reactant.

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

the balanced equation, we know that the molar ratio of carbon to hydrogen is

1:2. Therefore, in the combustion of the compound CaHb,amoles of carbon

will combine with 2amoles of hydrogen.

Given that moles of H2O=0.12775 mol, we have 0.12775 mol = 2a→a=

0.063875.

Thus, the compound has 0.063875 moles of carbon and 2 ×0.063875 =

0.12775 moles of hydrogen.

Step 4: Calculate the molar ratio of carbon to hydrogen. The molar ratio of

carbon to hydrogen in the compound is:

0.063875 mol C

0.12775 mol H =1

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

formula of the compound is C1H2.

Question 30

Question

A gaseous compound containing only carbon and hydrogen was found to have

the following mass percents: 85.6

Solution

Step 1: Assume we have 100 g of the compound. This means that 85.6 g is

carbon and 14.4 g is hydrogen.

Step 2: Calculate the number of moles of each element using their respective

molar masses. For carbon:

moles of carbon = 85.6 g

12.01 g/mol = 7.13 mol

For hydrogen:

moles of hydrogen = 14.4 g

1.01 g/mol = 14.26 mol

Step 3: Determine the empirical formula of the compound by finding the

simplest whole number ratio of carbon to hydrogen. To find the ratio, we divide

the moles by the smallest number of moles (7.13):

Carbon = 7.13 mol

7.13 mol = 1

Hydrogen = 14.26 mol

7.13 mol = 2

Therefore, the empirical formula is CH2.

Step 4: Calculate the molar mass of the empirical formula:

Molar mass of CH2= 12.01 g/mol + 2(1.01 g/mol) = 14.03 g/mol

Step 5: Determine the molecular formula by dividing the given molar mass

by the molar mass of the empirical formula.

Molecular formula = 78.11 g/mol

14.03 g/mol = 5.56

Since 5.56 is close to 6, the molecular formula is (CH2)6, which simplifies to

C6H12.

Question 31

Question

A gaseous compound composed of nitrogen and oxygen has a volume of 2.00 L

at 27

C and 1.00 atm. When this compound is heated to 927

C, the volume

expands to 3.50 L at 1.00 atm. If the compound contains 70

(Assume ideal gas behavior and that the gases do not react with each other.)

Solution

Step 1: Calculate the initial number of moles of the compound at 27

C and

1.00 atm using the ideal gas law. Given: Initial volume, V1= 2.00 L Initial

temperature, T1= 27C= 300 K Initial pressure, P1= 1.00 atm

Using the ideal gas law P V =nRT , where R= 0.0821 L

atm/(mol

K), we

can solve for the initial number of moles, n1:

n1=P1V1

RT1

n1=(1.00 atm)(2.00 L)

(0.0821 L

atm/(mol

K))(300 K)

n1= 0.083mol

Step 2: Calculate the final number of moles of the compound at 927

C and

1.00 atm. Given: Final volume, V2= 3.50 L Final temperature, T2= 927C=

1200 K Final pressure, P2= 1.00 atm

Using the ideal gas law, we can solve for the final number of moles, n2:

n2=P2V2

RT2

n2=(1.00 atm)(3.50 L)

(0.0821 L

atm/(mol

K))(1200 K)

n2= 0.343mol

Step 3: Calculate the number of moles of nitrogen (N2) and oxygen (O2) in

the compound. Given: Percentage of nitrogen, 70

From the initial number of moles calculated in Step 1: - Moles of N2=

0.70(0.083mol)=0.058mol - Moles of O2= 0.083mol −0.058mol = 0.025mol

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

moles of nitrogen to oxygen is:

0.058 mol

0.025 mol =2.32

1≈2:1

Therefore, the empirical formula of the compound is N2O.

Question 32

Question

A gaseous compound containing only carbon and hydrogen is burned in oxygen

gas. The combustion of 5.00 g of the compound produced 12.0 g of carbon

dioxide and 3.70 g of water vapor. Determine the molecular formula of the

compound.

Solution

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

pound:

Compound + O2→CO2+ H2O

Since we are given the amounts of CO2and H2O produced, we will have to find

the amount of the compound that reacts with that oxygen to produce them.

Step 2: Calculate the moles of CO2produced:

Moles of CO2=Mass of CO2

Molar mass of CO2

=12.0 g

44.01 g/mol = 0.272 mol

Step 3: Calculate the moles of H2O vapor produced:

Moles of H2O = Mass of H2O

Molar mass of H2O=3.70 g

18.02 g/mol = 0.205 mol

Step 4: Determine the limiting reagent by comparing the moles of CO2and

H2O produced. Since CO2is produced from the carbon in the compound, and

H2O is produced from the hydrogen in the compound, we need to find which of

the two reactants will completely react.

Step 5: Calculate the moles of the compound that reacted based on the

moles of CO2produced, using the balanced chemical equation:

Moles of compound = Moles of CO2

1= 0.272 mol

Step 6: Calculate the mass of the compound that reacted:

Mass of compound = Moles of compound ×Molar mass of the compound

Step 7: Determine the molecular formula of the compound by finding the

ratio of the actual mass of the compound to the molar mass of the empirical

formula.

Question 33

Question

A 5.00 L container holds gas at a pressure of 1.50 atm and a temperature of

273 K. If the gas is released until the pressure inside the container reaches 1.00

atm, what will be the new volume of the gas at the same temperature?

Solution

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

P V =nRT . Given: P1= 1.50 atm V1= 5.00 L T= 273 K R= 0.0821 L atm

/ (mol K)

We first convert the pressure to atm: P1= 1.50 atm

Now we rearrange the ideal gas law equation to solve for n:n=P1V1

Substitute the known values: n=(1.50 atm)·(5.00 L)

(0.0821 L atm / (mol K))·(273 K)

n=7.50

22.4153 ≈0.335 mol

So, the initial number of moles of gas is approximately 0.335 mol.

Step 2: Use the equation P V =nRT to find the new volume when the

pressure decreases to 1.00 atm. Now, we rearrange the ideal gas law equation

to solve for V2with the new pressure P2:V2=nRT

Substitute the known values: V2=(0.335 mol)·(0.0821 L atm / (mol K))·(273 K)

1.00 atm

V2=7.50

1.00 ≈7.50 L

Therefore, the new volume of the gas when the pressure reaches 1.00 atm is

approximately 7.50 L.

Question 34

Question

When 8.00 L of hydrogen gas reacts with excess nitrogen gas at STP, ammonia

gas is formed. Calculate the volume of ammonia gas produced at STP. The

balanced chemical equation is:

3H2(g) + N2(g)→2NH3(g)

Solution

Step 1: Determine the moles of hydrogen gas given the volume and conditions.

Given:



Volume of hydrogen gas (VH2) = 8.00 L



Conditions: Standard Temperature and Pressure (STP), which is 0

C (273

K) and 1 atm pressure

Using the ideal gas law:

P V =nRT

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

constant (0.08206 L atm/mol K) T= temperature (in K)

Since the gas is at STP, we know that P= 1 atm and T= 273 K.

Plugging in the values, we get:

1×8.00 = n×0.08206 ×273

8.00 = 22.3778n

n= 0.3571 moles

Step 2: Determine the volume of ammonia gas produced by using mole ratio

from the balanced chemical equation. From the balanced equation:

3 moles of H2: 2 moles of NH3

0.3571 moles of H2:xmoles of NH3

Solving for x, we get:

x=0.3571 ×2

3= 0.2381 moles of NH3

Step 3: Calculate the volume of ammonia gas at STP. Using the ideal gas

law again:

P V =nRT

For ammonia gas:

P= 1 atm

T= 273 K

Plugging in n= 0.2381 and R= 0.08206, we get:

1×VNH3= 0.2381 ×0.08206 ×273

VNH3=0.2381 ×0.08206 ×273

1= 4.88 L

Therefore, the volume of ammonia gas produced at STP is 4.88 L.

Question 35

Question

A certain hydrocarbon reacts with oxygen gas to produce carbon dioxide and

water vapor. If 2.00 L of the hydrocarbon at STP completely reacts with excess

oxygen, what volume of carbon dioxide is produced at STP?

Solution

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

chemical equation for the combustion of a hydrocarbon with oxygen is:

Hydrocarbon + O2→CO2+ H2O

Step 2: Determine the molar ratio between the hydrocarbon and carbon

dioxide. From the balanced chemical equation, we can see that for every one

mole of hydrocarbon reacted, one mole of carbon dioxide is produced.

Step 3: Calculate the number of moles of the hydrocarbon. Since the gas is

at STP, we can use the ideal gas law to find the number of moles:

P V =nRT

n=P V

Plugging in the values (P= 1.00 atm, V= 2.00 L, R= 0.0821 L atm/mol K,

T= 273 K):

n=(1.00 atm)(2.00 L)

(0.0821 L atm/mol K)(273 K) ≈0.0878 mol

Step 4: Use the molar ratio to find the number of moles of carbon dioxide

produced. Since the molar ratio between the hydrocarbon and carbon dioxide

is 1:1, the number of moles of carbon dioxide produced is also approximately

0.0878 mol.

Step 5: Calculate the volume of carbon dioxide produced at STP. Now, we

can use the ideal gas law again to calculate the volume of carbon dioxide at

STP:

P V =nRT

Solution

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

and oxygen: C3H8+ 5O2→3CO2+ 4H2O

Step 2: Determine the mole ratio between propane and carbon dioxide from

the balanced chemical equation. 1 mol of propane produces 3 mol of carbon

dioxide.

Step 3: Calculate the number of moles of propane present:

Volume of propane given = 10.0 L

From the ideal gas law, we know that:

P V =nRT

n=P V

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

gas constant, and Tis the temperature. Since the temperature and pressure are

constant in this case, we can simplify the formula to:

n=V

22.4

n=10.0 L

22.4 L/mol = 0.4464 mol

Step 4: Use the mole ratio to calculate the number of moles of carbon dioxide

produced:

0.4464 mol propane ×3 mol CO2

1 mol propane = 1.3392 mol CO2

Step 5: Convert the number of moles of carbon dioxide to volume using the

ideal gas law:

n=P V

V=nRT

V=1.3392 mol ×0.0821 (atm ·L/mol ·K) ×273 K

1 atm = 30.3 L

Therefore, 30.3 liters of carbon dioxide gas will be produced.

Question 3

Question

In a chemical reaction, 3.00 L of hydrogen gas reacts with excess nitrogen gas

to produce 2.00 L of ammonia gas at the same temperature and pressure. The

balanced chemical equation for the reaction is:

3H2(g) + N2(g)→2NH3(g)

Determine the volume of hydrogen gas (in L) that would be needed to pro-

duce 10.0 L of ammonia gas under the same conditions.

Solution

Step 1: Calculate the molar ratio of hydrogen gas to ammonia gas based on the

balanced chemical equation. Step 2: Use the molar ratio to find the volume of

hydrogen gas needed to produce 10.0 L of ammonia gas.

Step 1: Calculate the molar ratio of hydrogen gas to ammonia gas.

From the balanced chemical equation, 3 moles of hydrogen gas react to

produce 2 moles of ammonia gas.

Step 2: Use the molar ratio to find the volume of hydrogen gas needed.

Given: Volume of ammonia gas produced = 10.0 L

Using the molar ratio from Step 1:

3 moles H2

2 moles NH3

=10.0 L H2

VL NH3

Solving for V:

V=10.0×2

3= 6.67 L

Therefore, 6.67 L of hydrogen gas would be needed to produce 10.0 L of

ammonia gas under the same conditions.

Question 4

Question

A reaction between hydrogen gas (H2) and nitrogen gas (N2) produces ammonia

gas (NH3) according to the equation:

3H2(g)+N2(g)→2NH3(g)

If 14.0 L of hydrogen gas reacts with an excess of nitrogen gas at STP, what

volume of ammonia gas is produced?

Solution

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

PV = nRT

Since the gases are at STP, the pressure (P) is 1 atm, the temperature (T) is

273 K, and the gas constant (R) is 0.0821 L atm/mol K.

n = PV

RT =(1 atm)(14.0 L)

(0.0821 L atm/mol K)(273 K) ≈0.787 mol

Step 2: Use the mole ratio from the balanced chemical equation to find the

moles of ammonia gas produced. From the balanced chemical equation, the

ratio of H2to NH3is 3:2.

nNH3=3

2×0.787 ≈1.18 mol

Step 3: Calculate the volume of ammonia gas produced using the ideal gas

law.

V = nRT

P=(1.18 mol)(0.0821 L atm/mol K)(273 K)

1 atm ≈27.8 L

Therefore, approximately 27.8 L of ammonia gas is produced when 14.0 L

of hydrogen gas reacts with an excess of nitrogen gas at STP.

Question 5

Question

A 2.50 L container at 25

C contains carbon monoxide gas at a pressure of 0.800

atm. If the carbon monoxide is burned to produce carbon dioxide according to

the following balanced chemical equation:

2CO(g) + O2(g)→2CO2(g)

What will be the pressure in the container after the reaction is complete

if the temperature remains constant and the volume of the container does not

change?

Solution

Step 1: Calculate the number of moles of carbon monoxide present in the con-

tainer using the ideal gas law equation:

P V =nRT

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

Ris the ideal gas constant, and Tis the temperature in Kelvin. Rearranging

the equation to solve for n:

n=P V

⇒n=(0.800 atm)(2.50 L)

(0.0821 atm ·L/mol ·K)(298 K)

⇒n≈0.082 mol

Step 2: Use the balanced chemical equation to determine the stoichiometry

between carbon monoxide and carbon dioxide. Since 2 moles of carbon monoxide

produce 2 moles of carbon dioxide, the mole ratio is 1:1.

Step 3: Calculate the number of moles of carbon dioxide produced:

moles of CO2= moles of CO

⇒moles of CO2= 0.082 mol

Step 4: Calculate the new pressure using the ideal gas law, assuming the

volume and temperature remain constant:

P2=nRT

⇒P2=(0.082 mol)(0.0821 atm ·L/mol ·K)(298 K)

2.50 L

⇒P2≈0.833 atm

Therefore, the pressure in the container after the reaction is complete will

be approximately 0.833 atm.

Question 6

Question

A mixture of ethane (C2H6) and oxygen (O2) is burned in a gas cylinder ac-

cording to the following balanced equation:

2C2H6(g) + 7O2(g)→4CO2(g) + 6H2O(g)

If 10.0 L of ethane at STP (standard temperature and pressure) and 5.00 L of

oxygen at STP are allowed to react, what volume in Liters of carbon dioxide at

STP will be produced?

Solution

Step 1: Calculate the number of moles for each reactant. Given that we have

10.0 L of ethane and 5.00 L of oxygen at STP, we can calculate the number

of moles for each using the ideal gas law, P V =nRT , where P= 1.00 atm,

V= 10.0 L (for ethane) and 5.00 L (for oxygen), T= 273 K and R= 0.0821 L ·

atm/mol ·K.

For ethane:

nC2H6=P V

RT =(1.00 atm)(10.0 L)

(0.0821 L ·atm/mol ·K)(273 K)

nC2H6=10.0

22.414 = 0.446 moles of C2H6

For oxygen:

nO2=P V

RT =(1.00 atm)(5.00 L)

(0.0821 L ·atm/mol ·K)(273 K)

nO2=5.00

22.414 = 0.223 moles of O2

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

tion, we see that the ratio of ethane to oxygen is 2:7. We can calculate the

moles of carbon dioxide produced by each reactant: Moles of CO2produced

from ethane: 0.446 moles ×4 moles CO2

2 moles C2H6= 0.892 moles CO2Moles of CO2pro-

duced from oxygen: 0.223 moles ×4 moles CO2

7 moles O2= 0.128 moles CO2

Since oxygen produces fewer moles of carbon dioxide, oxygen is the limiting

reactant.

Step 3: Calculate the volume of carbon dioxide produced. Since the reaction

ratio shows that 4 moles of CO2 are produced for every 7 moles of O2, we can

calculate the volume of CO2 produced using the ideal gas law:

VCO2=nRT

P=(0.128)(0.0821)(273)

1.00 = 2.66 L

Therefore, 2.66 Liters of carbon dioxide at STP will be produced.

Question 7

Question

A reaction takes place between 3.00 L of oxygen gas at STP and excess hydrogen

gas to produce water vapor. If the reaction yields 4.50 L of water vapor at STP,

what is the balanced chemical equation for the reaction?

Solution

Step 1: Determine the molar ratio between oxygen and water vapor based on

the given volumes at STP.

Given: - Volume of oxygen gas = 3.00 L - Volume of water vapor produced

= 4.50 L - Both volumes are at STP

From the ideal gas law, we know that 1 mole of any ideal gas at STP occupies

22.4 L. Therefore: - 3.00 L of oxygen gas at STP = 3.00 L / 22.4 L/mol = 0.134

mols of O2 - 4.50 L of water vapor at STP = 4.50 L / 22.4 L/mol = 0.201 mols

of H2O

Step 2: Determine the molar ratios of oxygen to water vapor in the balanced

chemical equation.

The reaction between oxygen and hydrogen to form water can be represented

as:

2H2+ O2→2H2O

From the balanced chemical equation, we see that 1 mol of O2 yields 2 mol

of H2O. Therefore, the molar ratio of O2 to H2O is 1:2.

Step 3: Write the balanced chemical equation for the reaction.

The balanced chemical equation is:

2H2+ O2→2H2O

Question 8

Question

A gaseous compound containing only carbon and hydrogen is combusted in

excess oxygen gas. If the combustion of 2.00 g of the compound produces 7.52

g of carbon dioxide and 2.59 g of water vapor, determine the empirical formula

of the compound.

Solution

Step 1: Find the number of moles of carbon dioxide produced. Given: mCO2=

7.52 g The molar mass of carbon dioxide (CO2) is 44.01 g/mol.

Number of moles of CO2=mCO2

Molar mass of CO2

=7.52 g

44.01 g/mol ≈0.171 mol

Step 2: Find the number of moles of water vapor produced. Given: mH2O=

2.59 g The molar mass of water (H2O) is 18.02 g/mol.

Number of moles of H2O = mH2O

Molar mass of H2O=2.59 g

18.02 g/mol ≈0.144 mol

Step 3: Determine the number of moles of carbon and hydrogen that combine

to form the compound. From the balanced combustion reaction equation, one

mole of the compound reacts to produce one mole of CO and one mole of

HO. Therefore, the compound produced 0.171 moles of carbon for every mole

of compound. Similarly, the compound produced 0.144 moles of hydrogen for

every mole of compound.

Step 4: Determine the empirical formula of the compound. Since the com-

pound contains only carbon and hydrogen, the empirical formula is CxHy. From

step 3, the compound has a ratio of 0.171 moles of carbon to 0.144 moles of hy-

drogen. To find the simplest whole number ratio, divide both values by the

smaller number of moles (0.144):

0.171 mol/0.144 mol

0.144 mol/0.144 mol =1.188

1≈6

This means the compound has an empirical formula of C6H5.

Question 9

Question

A gaseous compound made up of nitrogen and chlorine has a molar volume of

50.8 L/mol at STP (0

C and 1 atm). If the compound contains 25

Solution

Step 1: Determine the masses of nitrogen and chlorine in 1 mole of the com-

pound. Let’s assume we have 100 g of the compound. - Mass of nitrogen in the

100 g compound: 25- Mass of chlorine in the 100 g compound: 100 g - 25 g =

75 g

Step 2: Convert the masses of nitrogen and chlorine to moles. - Moles of

nitrogen: 25g

14.01g/mol ≈1.78 mol (using atomic mass of nitrogen = 14.01 g/mol) -

Moles of chlorine: 75g

35.45g/mol ≈2.11 mol (using atomic mass of chlorine = 35.45

g/mol)

Step 3: Determine the simplest whole number ratio of moles of nitrogen to

moles of chlorine. By dividing the moles of each element by the smaller number

of moles: - For nitrogen: 1.78mol

1.78mol ≈1 - For chlorine: 2.11mol

1.78mol ≈1.19

Step 4: Multiply the ratio by a whole number to get whole numbers for the

subscripts in the empirical formula. Since the ratio for chlorine is close to 1.19,

we can round it off: - For nitrogen: 1 ×1 = 1 - For chlorine: 1.19 ×1≈1.19 ⇒

round to 1

Therefore, the empirical formula of the compound is NCl.

Question 10

Question

A reaction between 25.0 g of methane (CH4) and 30.0 g of oxygen gas (O2)

produces carbon dioxide (CO2) and water (H2O) according to the following

balanced chemical equation:

CH4(g) + 2O2(g)→CO2(g) + 2H2O(g)

What is the limiting reactant in this reaction? How many grams of water

are produced?

Solution

Step 1: Calculate the number of moles of each reactant.



Moles of methane (CH4):

Moles = Mass

Molar Mass =25.0 g

16.05 g/mol ≈1.556 mol



Moles of oxygen gas (O2):

Moles = Mass

Molar Mass =30.0 g

32.00 g/mol ≈0.938 mol

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

we compare the moles of each reactant to the stoichiometry of the balanced

equation.



For methane: 1.556 mol CH4

1= 1.556 mol O2need for complete reaction.



For oxygen gas: 0.938 mol O2

2= 0.469 mol CH4need for complete reaction.

Since we need 1.556 mol of O2but only have 0.938 mol, oxygen gas is the

limiting reactant.

Step 3: Calculate the mass of water produced using the limiting reactant.



Moles of water produced:

Moles = Moles of limiting reactant×Coefficient of water

Coefficient of limiting reactant = 0.938 mol×2

2= 0.938 mol



Mass of water produced:

Mass = Moles ×Molar Mass = 0.938 mol ×18.015 g/mol = 16.93 g

Therefore, the limiting reactant is oxygen gas and 16.93 g of water are pro-

duced.

Question 11

Question

A sample of a gaseous compound containing only carbon and hydrogen was

burned in excess oxygen. The volume of carbon dioxide produced was found to

be 2.50 L at 25

C and 1 atm. If the initial mass of the compound was 3.00 g,

what is the empirical formula of the compound?

(Take the molar volume of a gas at STP to be 22.4 L/mol, and the molar

mass of carbon dioxide to be 44.01 g/mol.)

Solution

Step 1: Calculate the number of moles of carbon dioxide produced. The volume

of carbon dioxide produced is 2.50 L. From the ideal gas law, we can calculate

the number of moles of carbon dioxide using the formula:

n=P V

Substitute P= 1 atm, V= 2.50 L, R= 0.0821 L·atm/mol·K, and T=

25C= 298 K into the formula:

n=(1 atm)(2.50 L)

(0.0821 L ·atm/mol ·K)(298 K)

n≈0.103 mol

Step 2: Calculate the number of moles of the compound. From the balanced

chemical equation, we know that the compound reacts to produce 1 mol of car-

bon dioxide. Since 1 mol of carbon dioxide is produced from 1 mol of compound,

the number of moles of the compound is also 0.103 mol.

Step 3: Calculate the molar mass of the compound. Given that the initial

mass of the compound is 3.00 g, and the number of moles of the compound is

0.103 mol, we can calculate the molar mass using the formula:

Molar mass = Total mass

Number of moles

Molar mass = 3.00 g

0.103 mol

Molar mass ≈29.13 g/mol

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

formula of the compound can be found by dividing the molar mass by the molar

mass of each element and then multiplying by the smallest whole number ratio.

To do this, we need to determine the number of moles of carbon and hydrogen

in the compound. - Carbon: Molar mass of carbon is about 12.01 g/mol, so the

number of moles of carbon in the compound is:

Moles of carbon = 12.01

29.13 ×0.103