CHEM 131 - ADVANCED GENERAL CHEMISTRY I Gas Stoichiometry Question Bank Set 2

CHEM 131 - ADVANCED GENERAL

CHEMISTRY I - Gas Stoichiometry

Question Bank - Set 2

Liberty University

Question 1

Question

A gas mixture contains oxygen and nitrogen in a 3:5 molar ratio. If the total

pressure of the gas mixture is 1.2 atm and the temperature is 300 K, determine

the partial pressure of each gas in the mixture.

Solution

Step 1: Determine the mole fraction of oxygen and nitrogen in the gas mixture.

Let xbe the number of moles of oxygen and ybe the number of moles of

nitrogen. Since the molar ratio of oxygen to nitrogen is 3:5, we have the following

relationship: x

y=3

Step 2: Find the total moles of gas in the mixture. Since the total pressure

is related to the total number of moles of gas by the ideal gas law, we have:

P V =nRT

(1.2 atm)(V)=(x+y)(0.0821 L ·atm/mol ·K)(300 K)

V=(x+y)(0.0821)(300)

1.2

Step 3: Express the partial pressures of oxygen and nitrogen in terms of

their mole fractions. The partial pressure of each gas is given by:

Poxygen = mole fraction of oxygen ×total pressure

Pnitrogen = mole fraction of nitrogen ×total pressure

Step 4: Calculate the mole fractions of oxygen and nitrogen. Since the mole

fraction of oxygen and nitrogen is given by x

x+yand y

x+yrespectively, we get:

x+y=3

x+y=5

Step 5: Substitute the mole fractions into the expressions for partial pres-

sures. Substitute the mole fractions into the expressions for partial pressures to

find:

Poxygen =3

8×1.2 atm

Pnitrogen =5

8×1.2 atm

Step 6: Calculate the partial pressures of oxygen and nitrogen. Finally, solve

for Poxygen and Pnitrogen to find their partial pressures in the gas mixture.

Poxygen = 0.45 atm

Pnitrogen = 0.75 atm

Therefore, the partial pressure of oxygen in the gas mixture is 0.45 atm and

the partial pressure of nitrogen is 0.75 atm.

Question 2

Question

When solid aluminum is placed in hydrochloric acid, the following reaction

occurs:

2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

If 2.00 grams of aluminum reacts with an excess of hydrochloric acid, what

volume of hydrogen gas is produced at 25

C and 1 atm?

Solution

Step 1: Calculate the number of moles of aluminum used. Given: Mass of

aluminum = 2.00 g, molar mass of aluminum = 26.98 g/mol

Moles of Al =2.00 g

26.98 g/mol ≈0.074mol

Step 2: Determine the limiting reagent. The balanced chemical equation

tells us that 2 moles of aluminum react with 6 moles of HCl to produce 3 moles

of hydrogen gas.

Thus, for 0.074 mol of Al, we would need 6

2×0.074 = 0.222 mol of HCl to

react with it.

In excess HCl, the reaction consumes all the aluminum so HCl is the limiting

reagent.

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

law.

Given: T= 25C= 298 K,Pressure = 1 atm, molar volume of a gas at STP

= 22.4 L/mol

Volume of H2= Moles of H2×Molar volume of gas at STP

= 3 ×0.074 ×22.4≈4.2 L

Therefore, the volume of hydrogen gas produced at 25

C and 1 atm is ap-

proximately 4.2 L.

Question 3

Question

A 2.00 L container is filled with acetylene gas (C2H2) at a pressure of 1.50 atm

and a temperature of 25

C. The acetylene is then burned in oxygen gas to form

carbon dioxide and water vapor according to the following balanced chemical

equation:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

If the final pressure in the container is 3.00 atm at the same temperature, what

mass of water vapor was produced?

(Note: The molar mass of C2H2is 26.04 g/mol and the molar mass of H2O

is 18.02 g/mol.)

Solution

Step 1: Calculate the moles of acetylene gas using the ideal gas law.

PV = nRT

n = PV

RT =(1.50 atm)(2.00 L)

0.0821 L ·atm/mol ·K·(25 + 273) K

n≈0.124 mol

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

tion, we can see that 2 moles of acetylene produce 2 moles of water. Since each

mole of acetylene produces 2 moles of water, only 0.062 moles of water will be

produced.

Step 3: Calculate the volume of water vapor at the final pressure using the

ideal gas law.

n = PV

V = nRT

P=(0.062 mol)(0.0821 L ·atm/mol ·K·(25 + 273) K)

3.00 atm

V≈0.341 L

Step 4: Convert the volume of water vapor to mass using the molar mass of

water.

Mass = n ×molar mass = (0.062 mol)(18.02 g/mol)

Mass = 1.12 g

Therefore, approximately 1.12 grams of water vapor were produced during

the reaction.

Question 4

Question

A reaction between solid lithium hydroxide and gaseous carbon dioxide produces

solid lithium carbonate and water vapor. If 5.00 g of lithium hydroxide reacts

with excess carbon dioxide to produce 2.67 g of water vapor, what is the limiting

reactant? What is the theoretical yield of lithium carbonate in grams?

Solution

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

hydroxide (LiOH) and carbon dioxide (CO2).

LiOH(s) + CO2(g)→Li2CO3(s)+H2O(g)

Step 2: Calculate the molar mass of lithium hydroxide (LiOH).

Molar mass of LiOH = Molar mass of Li + Molar mass of O + Molar mass of H

Molar mass of LiOH = 6.94 g/mol + 15.999 g/mol + 1.008 g/mol = 23.948 g/mol

Step 3: Calculate the moles of water vapor produced.

Moles of H2O = Mass of H2O

Molar mass of H2O=2.67 g

18.015 g/mol

Moles of H2O≈0.148 mol

Step 4: Identify the stoichiometry of the reaction from the chemical equation.

From the balanced chemical equation, we see that 1 mole of lithium hydroxide

reacts with 1 mole of carbon dioxide to produce 1 mole of water vapor.

Step 5: Use the stoichiometry of the reaction to calculate the moles of lithium

hydroxide reacted. Since 1 mole of lithium hydroxide produces 1 mole of water

vapor, the moles of lithium hydroxide reacted are also approximately 0.148 mol.

Step 6: Calculate the theoretical yield of lithium carbonate. From the bal-

anced chemical equation, the molar ratio between lithium hydroxide and lithium

carbonate is 2:1.

Moles of Li2CO3= 0.148 mol ×1 mol Li2CO3

2 mol LiOH = 0.074 mol

Theoretical yield of Li2CO3= 0.074 mol ×Molar mass of Li2CO3

Step 7: Calculate the theoretical yield of lithium carbonate in grams.

Theoretical yield of Li2CO3= 0.074 mol ×73.89 g/mol = 5.47 g

Step 8: Conclusion The limiting reactant in the reaction is lithium hydroxide,

and the theoretical yield of lithium carbonate is 5.47 grams.

Question 5

Question

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

dioxide and water vapor. If 2.00 L of the hydrocarbon at STP (standard tem-

perature and pressure) produces 9.00 L of carbon dioxide, what is the molecular

formula of the hydrocarbon?

Solution

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

CxHy+ (x + y/4)O2−−→ xCO2+ (y/2)H2O

Step 2: Calculate the number of moles of carbon dioxide produced. Given:

Volume of hydrocarbon = 2.00 L Volume of carbon dioxide produced = 9.00 L

Since we are at STP (standard temperature and pressure), we can use the

volume ratios as mole ratios. Moles of carbon dioxide = Volume of carbon

dioxide / 22.4 L Moles of carbon dioxide = 9.00 L / 22.4 L = 0.40179 mol

Step 3: Determine the mole ratio of carbon dioxide to the hydrocarbon.

From the balanced equation, the mole ratio between carbon dioxide and the

hydrocarbon is 1:1.

Step 4: Determine the number of moles of the hydrocarbon. Since the mole

ratio between carbon dioxide and the hydrocarbon is 1:1, the number of moles

of the hydrocarbon is the same as the number of moles of carbon dioxide. Moles

of hydrocarbon = 0.40179 mol

Step 5: Calculate the molar mass of the hydrocarbon.

Molar mass of hydrocarbon = Mass of hydrocarbon

Number of moles

=Volume ×Density

Number of moles

=2.00 ×1.18

0.40179

= 5.91 g/mol

Step 6: Determine the molecular formula of the hydrocarbon. The molar

mass of the hydrocarbon is 5.91 g/mol. Since we know the molar mass of carbon

is 12.01 g/mol and the molar mass of hydrogen is 1.008 g/mol, we can find the

molecular formula. Dividing the molar mass of the hydrocarbon by the molar

mass of carbon (12.01 g/mol) gives the number of carbon atoms in the molecular

formula, and dividing by the molar mass of hydrogen (1.008 g/mol) gives the

number of hydrogen atoms.

Number of carbon atoms = 5.91

12.01 ≈0.49 (round to 1)

Number of hydrogen atoms = 5.91

1.008 ≈5.86 (round to 6)

Therefore, the molecular formula of the hydrocarbon is CH6.

Question 6

Question

A gas mixture contains O2and N2in a ratio of 1:3 by volume. If the total

pressure of the mixture is 2 atm and the temperature is 300 K, what is the

partial pressure of each gas in the mixture?

Solution

Step 1: Determine the mole fractions of each gas in the mixture. Let the volume

of O2be VO2and the volume of N2be VN2. Since the ratio of the volumes of

O2and N2is 1:3, we have VO2=Vand VN2= 3V.

Step 2: Calculate the moles of each gas. The moles of each gas can be cal-

culated using the ideal gas law: P V =nRT . Since the volume and temperature

are the same for both gases, the moles of each gas are directly proportional to

their partial pressures. Let the moles of O2be nO2and the moles of N2be nN2.

Step 3: Use the mole fractions to determine the partial pressures. The

total pressure in the mixture is the sum of the partial pressures of each gas:

Ptotal =PO2+PN2. Since the moles of each gas are directly proportional to their

partial pressures, the partial pressures can be expressed as PO2=nO2

ntotal

·Ptotal

and PN2=nN2

ntotal

·Ptotal.

Step 4: Substitute the mole fractions into the equations for partial pressures.

Since the mole fractions of O2and N2are 1/4 and 3/4 respectively, the partial

pressures can be calculated as follows: PO2=1/4

1/4+3/4·2 atm and PN2=3/4

1/4+3/4·

2 atm. Simplifying these expressions gives the partial pressures of each gas in

the mixture.

Question 7

Question

A container holds 5.00 L of pure oxygen gas at 0.900 atm and 20.0

C. A spark is

introduced, causing the oxygen to react with hydrogen gas to form water vapor.

If hydrogen gas is in excess and the reaction goes to completion, what volume

of water vapor is produced at 1.00 atm and 30.0

(Note: The balanced chemical equation for the reaction is 2H(g) + O(g)



2HO(g))

Solution

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

Volume of oxygen gas (V) = 5.00 L

Pressure of oxygen gas (P) = 0.900 atm

Temperature of oxygen gas (T) = 20.0C = 293.15 K

Pressure of water vapor = 1.00 atm

Temperature of water vapor = 30.0C = 303.15 K

Balanced Equation: 2H(g) + O(g)→2HO(g)

Step 2: Convert the initial conditions for oxygen gas to standard conditions.

P1V1=nRT1

(0.900 atm)(5.00 L) = n(0.0821 atm ·L/mol ·K)(293.15 K)

n=(0.900 atm)(5.00 L)

0.0821 atm ·L/mol ·K·293.15 K

n≈0.181 mol

Step 3: Use the mole ratio from the balanced equation to determine the

moles of water vapor produced.

Moles of HO produced = 2 ×nO2

= 2 ×0.181 mol

= 0.362 mol

Step 4: Convert the moles of water vapor produced to the final conditions.

P2V2=nRT2

(1.00 atm)(V2) = (0.362 mol)(0.0821 atm ·L/mol ·K)(303.15 K)

V2=(0.362 mol)(0.0821 atm ·L/mol ·K)(303.15 K)

1.00 atm

V2≈8.74 L

Therefore, the volume of water vapor produced at 1.00 atm and 30.0

C is

approximately 8.74 L.

Question 8

Question

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

oxygen. The combustion of 2.50 grams of this compound produced 9.00 grams

of carbon dioxide and 3.75 grams of water. Determine the molecular formula of

the compound.

Solution

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

compound. The combustion of a compound containing only carbon and hydro-

gen can be represented by the following equation:

CxHy +O2→CO2+H2O

Step 2: Calculate the moles of carbon dioxide produced. Given that 9.00

grams of carbon dioxide were produced, we first need to calculate the moles of

carbon dioxide:

Moles of CO2=9.00 g

44.01 g/mol = 0.204moles

Step 3: Calculate the moles of water produced. Given that 3.75 grams of

water were produced, we need to calculate the moles of water:

Moles of H2O=3.75 g

18.02 g/mol = 0.208moles

Step 4: Determine the moles of carbon and hydrogen in the compound. From

the balanced equation, we can see that for each mole of carbon dioxide produced,

one mole of carbon is present, and for each mole of water produced, two moles of

hydrogen are present. Therefore, the moles of carbon in the compound is equal

to the moles of carbon dioxide, and the moles of hydrogen in the compound is

twice the moles of water.

Moles of C = 0.204 moles Moles of H = 2 * 0.208 moles = 0.416 moles

Step 5: Find the molar ratio of carbon to hydrogen. To find the simplest

ratio of carbon to hydrogen, we divide both moles by the smallest value, in this

case 0.204: Moles of C = 0.204 moles

0.204 = 1 Moles of H = 0.416 moles

0.204 ≈2

Thus, the empirical formula of the compound is CH2.

Step 6: Calculate the molar mass of the empirical formula. The molar mass

of CH2is 12.01 g/mol + 2 ×1.01 g/mol = 14.03 g/mol.

Step 7: Find the molecular formula of the compound. Given that the molar

mass of the compound is 14.03 g/mol and the given mass of 2.50 grams, we

calculate the ratio of the molar mass of the compound to the molar mass of the

empirical formula:

Ratio = 2.50 g

14.03 g/mol = 0.178 mol

Since the ratio is approximately 1, the molecular formula of the compound

is the same as the empirical formula, CH2.

Question 9

Question

A gaseous compound contains only nitrogen, oxygen, and sulfur. It is found

that 0.500 L of the gas at 25

C and 1.00 atm pressure weighs 1.31 g. The gas

is then reacted with excess hydrogen gas to produce ammonia. If 0.250 L of

ammonia gas at the same temperature and pressure weighs 0.900 g, what is the

molecular formula of the gaseous compound?

(Note: Assume ideal gas behavior.)

Solution

Step 1: Calculate the molar mass of the gaseous compound.

Let mbe the mass of the gaseous compound and Mbe its molar mass. The

number of moles of the compound in 0.500 L volume at 25

C and 1.00 atm is

given by the ideal gas equation:

n=P V

where Pis the pressure, Vis the volume, Ris the ideal gas constant, and

Tis the temperature in Kelvin. Substituting the values:

n=(1.00 atm)(0.500 L)

(0.0821 L ·atm/K ·mol)(298 K) =0.500

24.48 = 0.02042 mol

Next, calculate the molar mass of the compound:

M=m

n=1.31 g

0.02042 mol ≈64.13 g/mol

Step 2: Calculate the number of moles of ammonia produced.

Using the same formula but with the new volume and mass:

nNH3=(1.00 atm)(0.250 L)

(0.0821 L ·atm/K ·mol)(298 K) =0.250

24.48 = 0.01021 mol

Step 3: Determine the ratio of moles of gaseous compound to moles of am-

monia.

Since the gaseous compound reacts with hydrogen to form ammonia, the stoi-

chiometry of the reaction must be considered. The balanced chemical equation

for the reaction is:

Compound + 3H2→NH3+ S

where S represents the unused sulfur from the compound. From the equation,

it can be seen that 1 mole of the gaseous compound produces 1 mole of ammonia.

Step 4: Calculate the molar mass ratio.

Since 1 mole of the compound produces 1 mole of ammonia, the ratio of their

molar masses is the same:

Mcompound

Mammonia

= 1

Therefore, the molecular formula of the gaseous compound must be the same

as that of ammonia, which is NH3.

Question 10

Question

Ammonia gas (NH3) is produced by the reaction of nitrogen gas (N2) with

hydrogen gas (H2) according to the following balanced chemical equation:

N2(g)+3H2(g)→2NH3(g)

If 5.00 L of nitrogen gas at STP and 3.00 L of hydrogen gas at STP are re-

acted according to the equation above, what volume of ammonia gas is produced

at STP?

Solution

Step 1: Determine the moles of nitrogen gas and hydrogen gas. Since the gases

are at STP (Standard Temperature and Pressure), we can use the molar volume

of gases at STP which is 22.4 L/mol.

Given: VN2= 5.00 L VH2= 3.00 L

Convert the volumes to moles:

nN2=VN2

22.4 L/mol =5.00 L

22.4 L/mol = 0.223 mol

nH2=VH2

22.4 L/mol =3.00 L

22.4 L/mol = 0.134 mol

Step 2: Determine the limiting reactant. The reaction uses 1 mole of nitrogen

gas for every 3 moles of hydrogen gas. Calculate the moles of ammonia gas

formed if all nitrogen gas reacts:

nNH3(if all N2reacts) = 2 ×nN2= 2 ×0.223 mol = 0.446 mol

Calculate the moles of ammonia gas formed if all hydrogen gas reacts:

nNH3(if all H2reacts) = 2

3×nH2=2

3×0.134 mol = 0.089 mol

Since the moles of ammonia gas formed is lower when all the hydrogen gas

reacts, hydrogen gas is the limiting reactant.

Step 3: Calculate the volume of ammonia gas produced. Using the volume-

mole relationship at STP:

VNH3=nNH3×22.4 L/mol = 0.089 mol ×22.4 L/mol = 1.99 L

Therefore, 1.99 L of ammonia gas is produced at STP.

Question 11

Question

A mixture of oxygen and acetylene is commonly used in welding. If 20.0 grams

of oxygen react with acetylene to produce carbon dioxide and water according

to the following balanced chemical equation:

2C2H2+ 5O2→4CO2+ 2H2O

How many grams of acetylene are needed to react with all the oxygen?

(Assume excess acetylene is present.)

Solution

Step 1: Calculate the moles of oxygen given the mass provided. Step 2: De-

termine the moles of acetylene needed based on the mole ratio between oxygen

and acetylene. Step 3: Convert moles of acetylene to grams.

Step 1: Given mass of oxygen: 20.0 grams Molar mass of oxygen (O2): 32.00

g/mol

Calculate moles of oxygen:

moles of O2=20.0 g

32.00 g/mol = 0.625 mol

Step 2: From the balanced chemical equation, the mole ratio of oxygen to

acetylene is 5:2. This means for every 5 moles of oxygen, 2 moles of acetylene

are needed.

Calculate moles of acetylene needed:

moles of C2H2=0.625 mol O2

5×2

1= 0.250 mol C2H2

Step 3: Molar mass of acetylene (C2H2): 26.04 g/mol

Convert moles of acetylene to grams:

mass of C2H2= 0.250 mol ×26.04 g/mol = 6.51 grams

Therefore, 6.51 grams of acetylene are needed to react with all the oxygen.

Question 12

Question

A mixture of gaseous hydrogen and gaseous oxygen is ignited and reacts to form

water vapor according to the following balanced chemical equation:

2H2(g)+O2(g)→2H2O(g)

If the initial volume of the mixture is 10.0 L at 20

C and 1.00 atm, what will

be the final volume of the water vapor formed at the same temperature and

pressure?

(Note: Assume all gases are ideal gases.)

Solution

Step 1: Calculate the moles of hydrogen and oxygen present in the initial mix-

ture using the ideal gas law.

Given: Initial volume (Vi) = 10.0 L Initial temperature (T) = 20

C = 293

K Initial pressure (P) = 1.00 atm

The ideal gas law is given by:

P V =nRT

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

ideal gas constant, and Tis the temperature in Kelvin.

Rearranging the formula to solve for moles:

n=P V

For hydrogen: nH2=(1.00 atm)(10.0 L)

(0.0821 L ·atm/K ·mol)(293 K)

nH2=10.0

(0.0821)(293) ≈0.42 mol

For oxygen: nO2=(1.00 atm)(10.0 L)

(0.0821 L ·atm/K ·mol)(293 K)

nO2=10.0

(0.0821)(293) ≈0.42 mol

Step 2: Determine the limiting reactant.

Since the balanced chemical equation shows that 2 moles of hydrogen react

with 1 mole of oxygen, we can see that each gas has the same number of moles

(0.42 mol). Therefore, oxygen is the limiting reactant.

Step 3: Calculate the moles of water vapor formed based on the limiting

reactant.

Since oxygen is the limiting reactant, it will react completely. From the

balanced chemical equation, 1 mole of oxygen forms 2 moles of water vapor.

nH2O= (0.42 mol O2)×2 mol H2O

1 mol O2

= 0.84 mol

Step 4: Calculate the final volume of water vapor using the ideal gas law.

Given: nH2O= 0.84 molVf=?

Using the ideal gas law:

V=nRT

Substitute the known values to solve for the final volume: Vf=(0.84 mol)(0.0821 L ·atm/K ·mol)(293 K)

1.00 atm

Vf=0.84 ×0.0821 ×293

1≈20.0 L

Therefore, the final volume of water vapor formed at the same temperature

and pressure would be approximately 20.0 L.

Question 13

Question

A mixture of hydrogen gas and oxygen gas undergoes a reaction to produce

water vapor according to the equation:

2H2(g) + O2(g)→2H2O(g)

If 4.25 L of hydrogen gas reacts with 2.13 L of oxygen gas at the same

temperature and pressure, what volume of water vapor is produced?

Solution

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

chemical equation is:

2H2(g) + O2(g)→2H2O(g)

Step 2: Determine the limiting reactant. Calculate the number of moles of

each reactant: For hydrogen gas (H2):

nH2=4.25 L

22.4 L/mol = 0.190 mol

For oxygen gas (O2):

nO2=2.13 L

22.4 L/mol = 0.095 mol

Because the balanced chemical equation has a 2:1 ratio of (H2) to (O2), it

is clear that there is not enough oxygen gas to react with all the hydrogen gas.

Therefore, oxygen gas is the limiting reactant.

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

chemical equation, we see that 1 mol of (O2) reacts to produce 2 mol of (H2O).

Therefore, the number of moles of (H2O) produced is:

nH2O= 0.095 mol ×2=0.190 mol

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

law equation:

P V =nRT

Since the temperature, pressure, and gas constant are constant:

V=n

×RT

Where: n= 0.190 mol (moles of water vapor produced) NA= 6.022 ×

1023mol−1(Avogadro’s number) R= 0.0821 atm ·L/mol ·K (gas constant) T

(temperature) and P(pressure) are assumed constant.

Plugging in the values:

V=0.190 mol

6.022 ×1023mol−1×0.0821 atm ·L/mol ·K×T

Since Tand Pare constant, we can simplify the equation to:

V=0.190

6.022 ×1023 ×0.0821 = 4.14 ×10−24 L

Therefore, 4.14 x 10−24 L of water vapor is produced.

Question 14

Question

A gaseous compound is composed of nitrogen and oxygen in a ratio of 2:1 by

volume. If 10.0 L of the compound reacts completely with hydrogen to form

ammonia gas and water vapor, calculate the volume of ammonia gas produced

at the same temperature and pressure.

N2(g)+3H2(g)→2NH3(g)

Solution

Step 1: Calculate the volume of nitrogen and oxygen in the compound. Let

the volume of nitrogen be 2xL and the volume of oxygen be xL, where xis a

common factor. Therefore, the total volume of the compound is 2x+x= 3xL.

Step 2: Calculate the volume of hydrogen required. From the balanced

chemical equation, 3 moles of hydrogen are required to react with 1 mole of

nitrogen. Since 1 mole of gas occupies 22.4 L at STP, 3 moles of hydrogen will

occupy 3 ×22.4 = 67.2 L.

Step 3: Calculate the volume of ammonia produced. From the balanced

chemical equation, 1 mole of nitrogen reacts to form 2 moles of ammonia. Since

10.0 L of the compound reacts, we can say 3x L of the compound is equivalent

to 10.0 L. Therefore, x=10.0

3L. So, the volume of ammonia produced is

2×10.0

3=20.0

3L.

Question 15

Question

A gaseous compound contains only nitrogen and oxygen. When 2.00 g of the

compound is completely decomposed, 0.50 g of nitrogen is obtained. What is

the empirical formula of the compound?

Given atomic masses: N = 14.00 g/mol, O = 16.00 g/mol.

Solution

Step 1: Determine the moles of nitrogen obtained. Given the mass of nitrogen

obtained is 0.50 g, we can convert this to moles using the molar mass of nitrogen.

Moles of nitrogen = Mass of nitrogen

Molar mass of nitrogen =0.50 g

14.00 g/mol

Step 2: Determine the moles of oxygen obtained. The total mass of the

compound is 2.00 g. So, the mass of oxygen obtained would be 2.00 g - 0.50 g

= 1.50 g.

Moles of oxygen = Mass of oxygen

Molar mass of oxygen =1.50 g

16.00 g/mol

Step 3: Find the mole ratio of nitrogen to oxygen. Dividing the moles of

nitrogen by the smaller number of moles (in this case, moles of nitrogen) gives

us the mole ratio.

Mole ratio of N to O = Moles of nitrogen

Moles of nitrogen :Moles of oxygen

Moles of nitrogen

Step 4: Determine the empirical formula of the compound. Using the mole

ratio found in the previous step, we can write the empirical formula of the

compound. Remember to multiply the subscripts by an integer if necessary to

obtain whole numbers.

Question 16

Question

A 2.00 L container at 25

C contains a mixture of 1.00 mol of H2, 1.00 mol of

N2, and 2.00 mol of NH3. If the reaction N2(g) + 3H2(g)→2NH3(g) reaches

completion, what will be the total pressure in the container if all the substances

are at 613 mmHg? Assume ideal gas behavior.

Solution

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

chemical equation is:

N2(g) + 3H2(g)→2NH3(g)

Step 2: Determine the moles of excess reactants and the limiting reactant.

Initially, there was 1.00 mol of N2, 1.00 mol of H2, and 2.00 mol of NH3. The

limiting reactant is the one with the smallest number of moles needed for the

reaction. In this case, N2is the limiting reactant because it provides only 1.00

mol, while the stoichiometry of the reaction requires 1.50 mol in order to fully

react with all of the H2.

Step 3: Calculate the moles of NH3produced. Since 1 mol of N2produces

2 mol of NH3, and N2is the limiting reactant, the moles of NH3produced will

be:

1.00 mol N2×2 mol NH3

1 mol N2

= 2.00 mol NH3

Step 4: Calculate the total moles of gas in the container after the reaction.

After the reaction, we have: - 0.00 mol of N2, - 1.00 mol of H2, - 0.00 mol of

NH3(all converted to NH3), - 2.00 mol of NH3produced. This gives us a total

of 3.00 mol of gas in the container.

Step 5: Calculate the total pressure in the container using the ideal gas law.

The ideal gas law is:

P V =nRT

P=nRT

Plugging in the values:

P=(3.00 mol)(0.0821 L ·atm/mol ·K)(298 K)

2.00 L

P=7.35 atm ·L

2.00 L

P≈3.67 atm

Therefore, the total pressure in the container after the reaction is approxi-

mately 3.67 atm.

Question 17

Question

When 5.00 L of methane gas (CH4) reacts with excess oxygen gas, how many

liters of carbon dioxide gas (CO2) are produced at STP (standard temperature

and pressure)? The balanced chemical equation for the reaction is:

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

Solution

Step 1: Write down the balanced chemical equation to determine the mole ratio

between CH4and CO2.

CH4(g)+2O2(g)→CO2(g)+2H2O(g)

The mole ratio between CH4and CO2is 1:1.

Step 2: Calculate the number of moles of CH4using the ideal gas law at

STP.

n=P V

RT =(1atm)(5.00L)

0.0821atm ·L/mol ·K·273K ≈0.19 mol

Step 3: Since the mole ratio between CH4and CO2is 1:1, the number of

moles of CO2produced is also approximately 0.19 mol.

Step 4: Calculate the volume of carbon dioxide gas at STP using the ideal

gas law.

V=nRT

P=(0.19 mol)(0.0821atm ·L/mol ·K)(273 K)

1 atm = 4.1 L

Therefore, when 5.00 L of methane gas reacts with excess oxygen gas, 4.1 L

of carbon dioxide gas are produced at STP.

Question 18

Question

A mixture of hydrogen gas and nitrogen gas reacts to produce ammonia accord-

ing to the following balanced equation:

3H2(g) + N2(g)→2NH3(g)

If 5.00 L of hydrogen gas at STP (standard temperature and pressure) is reacted

with excess nitrogen gas, what volume of ammonia gas is produced?

Solution

Step 1: Determine the moles of hydrogen gas. Since the volume of hydrogen

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

of hydrogen gas:

P V =nRT

n=P V

n=(1.00 atm)(5.00 L)

(0.0821 L·atm/mol ·K)(273 K)

n= 0.223 mol H2

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

moles of ammonia. From the balanced chemical equation, we see that 3 moles

of hydrogen gas will produce 2 moles of ammonia.

0.223 molH2

3=x molN H3

x=0.223 mol

3×2

x= 0.149 mol NH3

Step 3: Calculate the volume of ammonia gas produced. Now we can use

the ideal gas law to find the volume of ammonia gas produced:

P V =nRT

V=nRT

V=(0.149 mol)(0.0821 L·atm/mol ·K)(273 K)

1.00 atm

V= 3.48 L NH3

Therefore, 3.48 L of ammonia gas is produced when 5.00 L of hydrogen gas

reacts with excess nitrogen gas.

Question 19

Question

A reaction takes place between aluminum metal and hydrochloric acid to pro-

duce aluminum chloride and hydrogen gas. If 10.0 grams of aluminum reacts

with an excess of hydrochloric acid, how many liters of hydrogen gas are pro-

duced at STP (Standard Temperature and Pressure)? (Assume complete reac-

tion and that the molar volume of a gas at STP is 22.4 L/mol)

Solution

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

chemical equation for the reaction between aluminum and hydrochloric acid is:

2Al + 6HCl →2AlCl3+ 3H2

Step 2: Calculate the number of moles of aluminum used. Given: Mass of

aluminum = 10.0 g, Molar mass of aluminum = 26.98 g/mol Number of moles

of aluminum = 10.0 g

26.98 g/mol

Step 3: Determine the limiting reactant to find the number of moles of

hydrogen gas produced. Since we have excess hydrochloric acid, aluminum is the

limiting reactant. From the balanced chemical equation, 2 moles of aluminum

produce 3 moles of hydrogen gas. So, moles of hydrogen gas produced = 3

2×

number of moles of aluminum

Step 4: Convert the number of moles of hydrogen gas to volume at STP.

Using the molar volume of a gas at STP (22.4 L/mol), we can convert the moles

of hydrogen gas to volume. Volume of hydrogen gas = moles of hydrogen gas

×molar volume at STP

Now, substitute the values into the equation and calculate the volume of

hydrogen gas produced in liters.

Question 20

Question

A gaseous compound containing only sulfur and fluorine is found to be 54.4

Solution

Step 1: Assume we have 100 g of the compound. This means we have 54.4 g of

sulfur and 45.6 g of fluorine.

Step 2: Calculate the number of moles of sulfur and fluorine. Number of

moles of sulfur:

moles of S = 54.4 g

32.06 g/mol = 1.696 mol

Number of moles of fluorine:

moles of F = 45.6 g

19.00 g/mol = 2.405 mol

Step 3: Determine the mole ratio by dividing by the smallest number of

moles.

Sulfur ratio = 1.696 mol

1.696 mol ≈1

Fluorine ratio = 2.405 mol

1.696 mol ≈1.42

Step 4: Find the empirical formula. Since we aim for whole numbers, we

will multiply the ratios by 2 to get the simplest whole number ratio:

S = 1 ×2 = 2 and F = 1.42 ×2 = 2.84

Step 5: Therefore, the empirical formula of the compound is SF2.

Question 21

Question

A mixture of hydrogen gas and oxygen gas is used in welding operations. If 15.0

L of hydrogen gas at STP reacts with oxygen gas, how many liters of oxygen gas

are required for the reaction to go to completion? Assume all gases are ideal.

Solution

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

gas and oxygen gas.

2 H2(g) + O2(g) −−→ 2 H2O(g)

Step 2: Determine the molar volume of a gas at STP. The molar volume of

a gas at STP (Standard Temperature and Pressure) is 22.4 L/mol.

Step 3: Calculate the number of moles of hydrogen gas using the ideal gas

law. Given: Volume of hydrogen gas, VH2 = 15.0 L Molar volume at STP,

Vmolar = 22.4 L/mol

We can use the formula: n=P V

RT where Pis the pressure, Vis the volume,

Ris the ideal gas constant, and Tis the temperature.

At STP, P= 1 atm and T= 273 K, so R= 0.0821 L atm/mol K.

Plugging in the values:

nH2 =(1 atm)(15.0 L)

(0.0821 L atm/mol K)(273 K) = 0.671 mol H2

Step 4: Use the stoichiometry of the balanced chemical equation to find the

quantity of oxygen gas required. From the balanced equation, we see that 2

moles of hydrogen gas react with 1 mole of oxygen gas.

Moles of O2=0.671 mol H2

2= 0.3355 mol O2

Step 5: Calculate the volume of oxygen gas at STP using the molar volume.

Volume of O2= 0.3355 mol ×22.4 L/mol = 7.52 L

Therefore, 7.52 liters of oxygen gas are required for the reaction to go to

completion.

Question 22

Question

In a reaction between hydrogen gas and oxygen gas, 10.0 L of hydrogen gas at

STP reacts with 5.00 L of oxygen gas at STP. If the reaction goes to completion,

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 (H2)

and oxygen gas (O2) to form water vapor (H2O) is:

2H2(g)+O2(g)→2H2O(g)

Step 2: Determine the moles of hydrogen gas and oxygen gas provided.

Since the gases are at STP, we can use the ideal gas law equation to calculate

the number of moles for hydrogen gas and oxygen gas:

PV = nRT

n = PV

where P is the pressure (1 atm), V is the volume in liters (10.0 L for hydrogen gas

and 5.00 L for oxygen gas), R is the ideal gas constant (0.0821 L

atm/mol

K),

and T is the temperature at STP (273 K).

For hydrogen gas:

nH2=(1 atm) ×(10.0 L)

(0.0821 L

atm/mol

K) ×(273 K)

nH2=10.0

22.414 ≈0.446 mol

For oxygen gas:

nO2=(1 atm) ×(5.00 L)

(0.0821 L

atm/mol

K) ×(273 K)

nO2=5.00

22.414 ≈0.223 mol

Step 3: Determine the limiting reactant.

From the balanced chemical equation, it is clear that hydrogen gas is the

limiting reactant since the molar ratio of hydrogen to oxygen is 2:1. We have

0.446 mol of hydrogen gas and 0.223 mol of oxygen gas.

Step 4: Calculate the volume of water vapor produced.

Since the reaction goes to completion, 2 moles of water vapor are produced

for every 2 moles of hydrogen gas consumed. Therefore, the number of moles of

water vapor produced is also 0.446 mol.

Now, we can use the ideal gas law to find the volume of water vapor:

V = nRT

P=(0.446 mol) ×(0.0821 L

atm/mol

K) ×(273 K)

1 atm

V≈10.0 L

Therefore, 10.0 L of water vapor is produced at STP.

Question 23

Question

A 2.00 L cylinder filled with oxygen gas at a pressure of 3.00 atm and a tem-

perature of 25

C. How many grams of potassium chlorate (KClO3) are needed

to produce oxygen gas at a pressure of 1.00 atm and a temperature of 50

Assume the reaction is given by:

2KClO3→2KCl + 3O2

Solution

Step 1: Find moles of oxygen gas present initially. Given: - Initial pressure of

oxygen gas (P1) = 3.00 atm - Initial volume of cylinder (V1) = 2.00 L - Initial

temperature of gas (T1) = 25

C = 298 K Using ideal gas law:

P V =nRT

n=P V

n=(3.00 atm)(2.00 L)

0.0821 L atm/mol K ·298 K

n= 0.243 mol

Step 2: Calculate the number of moles of oxygen gas needed to reach the

final conditions. Given: - Final pressure of oxygen gas (P2) = 1.00 atm - Final

temperature of gas (T2) = 50

C = 323 K From the ideal gas law:

n=P V

n=(1.00 atm)(2.00 L)

0.0821 L atm/mol K ·323 K

n= 0.062 mol

Step 3: Determine the number of moles of oxygen gas produced by the

reaction. Since the balanced chemical equation states that 2 moles of KClO3

produce 3 moles of O2, we can set up a ratio:

0.062 mol O2

3=xmol KClO3

Solving for x gives:

x=0.062 mol O2×2

x= 0.041 mol KClO3

Step 4: Determine the mass of potassium chlorate required. Given: - Molar

mass of KClO3= 122.55 g/mol Mass of KClO3needed:

Mass = moles ×Molar mass

Mass = 0.041 mol ×122.55 g/mol

Mass = 5.03 g

Therefore, 5.03 grams of potassium chlorate are needed to produce the spec-

ified amount of oxygen gas.

Question 24

Question

A 2.0 L container is filled with fluorine gas at a pressure of 1.0 atm and a

temperature of 25 degrees Celsius. If the container is heated to 125 degrees

Celsius, what will the pressure be if the volume is kept constant?

Solution

Step 1: Use the ideal gas law, P V =nRT , where Pis pressure, Vis volume, n

is the number of moles, Ris the ideal gas constant, and Tis temperature (in

Kelvin).

Step 2: Rearrange the ideal gas law to solve for P:

P=nRT

Step 3: Since the volume is constant, we can set the initial pressure and

temperature equal to the final pressure and temperature:

nR(273 + 25)

2.0=nR(273 + 125)

2.0

Step 4: Cancel out like terms:

298n= 398n

Step 5: Solve for the final pressure, Pf:

Pi=Pf

nRTi

V=nRTf

nR(273 + 25)

2.0=nR(273 + 125)

2.0

nR(298)

2.0=nR(398)

2.0

(298) = (398)

Pf= 1.0 atm

Step 6: Therefore, the pressure in the container when heated to 125 degrees

Celsius will still be 1.0 atm.

Question 25

Question

A 2.50 L container at 27

C initially contains 0.150 mol of CO and 0.190 mol of

CO2. The following reaction takes place:

2CO(g)+O2(g)→2CO2(g)

If the reaction goes to completion, what are the final amounts in moles of each

gas at 27

Solution

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

2CO(g)+O2(g)→2CO2(g)

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

we need to compare the amount of product that can be produced from each

reactant. We start with the moles of CO and CO2given. From the balanced

chemical equation, we can see that 2 moles of CO produce 2 moles of CO2.

Moles of CO2produced from CO = (0.150 mol CO) ×2 mol CO2

2 mol CO 

= 0.150 mol CO2

Next, from the coefficients in the balanced chemical equation, 2 moles of

CO2are produced from 1 mole of O2.

Moles of CO2produced from O2= (0.190 mol CO2)×2 mol CO2

1 mol O2

= 0.380 mol CO2

Since the amount of CO produces fewer moles of CO2compared to the

amount of O2, CO is the limiting reactant.

Step 3: Calculate the final amounts of each gas in moles. Since CO is

the limiting reactant, all of it will be consumed. From the balanced chemical

equation, 2 moles of CO produce 2 moles of CO2.

Amount of CO2= (0.150 mol CO) ×2 mol CO2

2 mol CO 

= 0.150 mol CO2

Since 0.150 mol of CO produces 0.150 mol of CO2, the final amounts of each

gas are: - CO: 0 mol - CO2: 0.150 mol - O2: 0.190 mol

Question 26

Question

A 2.00 L container at 303 K is filled with 3.00 atm of nitrogen gas and 2.00

atm of oxygen gas. The gases react to form nitrogen dioxide according to the

following chemical equation:

2 N2(g) + 4 O2(g)→4 NO2(g)

If the reaction goes to completion, what is the total pressure in the container

at the end? Assume all gases behave ideally.

Solution

Step 1: Calculate the moles of each gas present using the ideal gas law formula,

P V =nRT , where Pis the pressure, Vis the volume, nis the number of moles,

Ris the ideal gas constant (0.0821 L atm/mol K), and Tis the temperature in

Kelvin. For nitrogen:

moles of N2=P V

RT =(3.00 atm)(2.00 L)

(0.0821 L atm/mol K)(303 K) ≈0.250 moles

For oxygen:

moles of O2=P V

RT =(2.00 atm)(2.00 L)

(0.0821 L atm/mol K)(303 K) ≈0.164 moles

Step 2: Determine the limiting reactant by comparing the mole ratios of N2

and O2in the balanced chemical equation. Since the balanced equation shows

that 2 moles of N2react with 4 moles of O2, the mole ratio is 2:4 or 1:2. Given

that we have approximately 0.25 moles of N2and 0.164 moles of O2, the O2is

the limiting reactant.

Step 3: Calculate the total moles of N2and O2molecules consumed during

the reaction. 1 mole of O2reacts with 0.5 moles of N2(from the balanced

chemical equation). Hence, the number of moles of N2consumed is 0.5×0.164 ≈

0.082.

Step 4: Determine the moles of NO2produced using stoichiometry. From the

balanced chemical equation, 2 moles of N2produce 4 moles of NO2. Therefore,

0.082 moles of N2will produce 0.082 ×4≈0.328 moles of NO2.

Step 5: Calculate the total moles of gas at the end using the ideal gas

law. Since 0.082 moles of N2and 0.164 moles of O2have been consumed, and

0.328 moles of NO2have been produced, the total moles of gas at the end is

0.250 −0.082 + 0.164 −0.082 + 0.328 = 0.678 moles.

Step 6: Calculate the total pressure in the container at the end using the

ideal gas law.

Ptotal =nRT

V=(0.678 moles)(0.0821 L atm/mol K)(303 K)

2.00 L ≈8.34 atm

Therefore, the total pressure in the container at the end of the reaction is

approximately 8.34 atm.

Question 27

Question

A mixture of 3.0 liters of methane gas (CH4) and 2.0 liters of oxygen gas (O2)

react to form carbon dioxide (CO2) and water (H2O) gas. If the reaction goes

to completion, what volume of water vapor is produced at the same temperature

and pressure?

Solution

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

chemical equation for the reaction of methane and oxygen to form carbon dioxide

and water is:

CH4+ 2O2→CO2+ 2H2O

Step 2: Determine the moles of each gas using the ideal gas law. The ideal

gas law is given by:

P V =nRT

Since the temperature and pressure are constant, the equation simplifies to:

V=P

For methane (CH4):

nCH4

VCH4

=PCH4

nCH4

3.0=PCH4

For oxygen (O2):

nO2

VO2

=PO2

nO2

2.0=PO2

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

tion, we see that it takes 1 mole of methane to produce 2 moles of water vapor.

Therefore, the number of moles of water vapor produced from the methane gas

is:

2×nCH4

Step 4: Calculate the volume of water vapor produced. Since the volume

of gases is directly proportional to the number of moles of gas, the volume of

water vapor produced is:

2×nCH4×VH2O

Substitute the expression for moles of methane and the volume of water

vapor into the formula to find the volume of water vapor produced:

2×3.0×PCH4

RT ×VH2O

Question 28

Question

A reaction of 5.00 L of methane gas (CH4) at STP with excess oxygen pro-

duces carbon dioxide and water vapor. Calculate the volume of carbon dioxide

produced at STP.

Solution

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

CH4(g) + 2O2(g)→CO2(g) + 2H2O(g)

Step 2: Determine the mole ratio between methane and carbon dioxide using

the balanced chemical equation: 1 mole of CH4produces 1 mole of CO2. Convert

5.00 L of CH4to moles:

5.00 L CH4×1 mol CH4

22.4 L CH4

= 0.223 mol CH4

Since 1 mol of CH4produces 1 mol of CO2, 0.223 mol of CH4will produce 0.223

mol of CO2.

Step 3: Convert the moles of carbon dioxide to volume at STP:

0.223 mol CO2×22.4 L/mol = 5.00 L CO2

Answer: The volume of carbon dioxide produced at STP is 5.00 L.

Question 29

Question

A 2.00 L container is filled with 1.50 atm of nitrogen gas and 2.00 atm of

hydrogen gas at 25

C. If the gases react to form ammonia, what is the pressure

of the vessel at 25

C after the reaction is complete? Assume the reaction goes

to completion.

Solution

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

gas and hydrogen gas to form ammonia.

N2(g)+3H2(g)→2NH3(g)

Step 2: Determine the number of moles of each gas present using the ideal

gas law, P V =nRT . For nitrogen gas:

nN2=PN2V

RT =(1.50 atm)(2.00 L)

0.0821 L

atm/mol

K×(25 + 273) K

nN2= 0.122 mol

For hydrogen gas:

nH2=PH2V

RT =(2.00 atm)(2.00 L)

0.0821 L

atm/mol

K×(25 + 273) K

nH2= 0.163 mol

Step 3: Determine the limiting reactant. Since nitrogen gas and hydrogen

gas are in a 1:3 ratio, nitrogen is the limiting reactant as there is less nitrogen

gas than needed for complete reaction.

Step 4: Calculate the moles of ammonia produced. From the balanced

chemical equation, 1 mol of nitrogen produces 2 mol of ammonia. Therefore,

0.122 mol of nitrogen gas will produce 2 ×0.122 = 0.244 mol of ammonia.

Step 5: Calculate the pressure of the vessel after the reaction is complete.

Total moles of gas after the reaction:

ntotal =nH2+nN2−2nNH3= 0.163 mol+0.122 mol−2(0.244 mol) = −0.203 mol

Using the ideal gas law with the total moles and volume of the container:

Pfinal =ntotalRT

V=(−0.203 mol)(0.0821 L

atm/mol

K)(25 + 273) K

2.00 L

Pfinal =−2.47 atm

Therefore, the pressure inside the container after the reaction is complete is

2.47 atm.

Question 30

Question

A gaseous compound containing only carbon and hydrogen was analyzed and

found to be 85.7

Solution

Step 1: Calculate the molar mass of carbon and hydrogen. Let’s assume we

have 100 g of the compound, which means we have 85.7 g of carbon and 14.3 g

of hydrogen. The molar mass of carbon (C) is 12 g/mol and the molar mass of

hydrogen (H) is 1 g/mol.

Step 2: Calculate the moles of carbon and hydrogen. The number of moles

of carbon can be calculated as:

moles of C = mass of C

molar mass of C =85.7 g

12 g/mol

Similarly, the number of moles of hydrogen can be calculated as:

moles of H = mass of H

molar mass of H =14.3 g

1 g/mol

Step 3: Determine the empirical formula of the compound. Next, we need

to find the ratio of moles of carbon to moles of hydrogen:

moles of C

moles of H =85.7/12

14.3/1

Step 4: Find the empirical formula. Now we need to normalize the ratio to

obtain whole numbers. The ratio found in the previous step was approximately

7.14 : 14.3 which simplifies to 1 : 2. Therefore, the empirical formula of the

compound is CH2.

Step 5: Find the molecular formula. Given that the molar mass of the

compound is 78 g/mol and the empirical formula mass is 14 g/mol, we can find

the molecular formula:

molar mass of molecular formula = n×empirical formula mass

where nis the number of empirical units in the molecular formula. Solving for

n=molar mass of molecular formula

empirical formula mass =78 g/mol

14 g/mol

Thus, n= 5.57, which means the molecular formula is approximately 5CH2, or

C5H10.

Question 31

Question

A reaction occurs between 2.00 L of chlorine gas at STP and excess sodium

bromide, producing sodium chloride and bromine gas. Calculate the volume of

bromine gas produced at STP.

Solution

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

chemical equation for the reaction is:

Cl2+ 2NaBr →2NaCl + Br2

Step 2: Use the ideal gas law to calculate the moles of Cl2. Given that the

volume of Cl2is 2.00 L and the conditions are STP, we can use the ideal gas

law to find the moles of Cl2:

P V =nRT

n=P V

n=(1.00 atm)(2.00 L)

(0.0821 L ·atm/mol ·K)(273 K)

n=2.00 atm ·L

22.414 L ·mol−1

n= 0.0891 mol

Step 3: Determine the volume of Br2produced at STP. From the balanced

chemical equation, we see that 1 mole of Cl2produces 1 mole of Br2. Using the

ideal gas law at STP conditions (0

C and 1 atm):

V=nRT

V=(0.0891 mol)(0.0821 L ·atm/mol ·K)(273 K)

1 atm

V=2.27 L ·atm

1 atm

V= 2.27 L

Therefore, the volume of bromine gas produced at STP is 2.27 L.

Question 32

Question

A gaseous compound is formed by the reaction of hydrogen gas and nitrogen

gas. If 3.00 L of hydrogen gas reacts with excess nitrogen gas to produce 10.0 L

of the compound at the same temperature and pressure, determine the empirical

formula of the compound.

Solution

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

gas and nitrogen gas.

3H2+ N2→Compound

Step 2: Determine the moles of hydrogen gas (H) involved in the reaction

using the ideal gas law, PV = nRT.

N = P V

RT =(1.00 atm)(3.00 L)

(0.0821 L ·atm/mol ·K)(273 K) ≈0.137 mol H2

Step 3: Since the reaction uses 3 moles of hydrogen gas, we need to use 0.137

mol to determine the amount of compound produced.

0.137 mol H2

3= 0.0457 mol compound

Step 4: Using the volume of the compound produced (10.0 L) and the ideal

gas law, determine the moles of the compound.

10.0 L = nRT

P=⇒n=(1.00 atm)(10.0 L)

0.0821 L ·atm/mol ·K= 1.22 mol

Step 5: Calculate the mole ratio of hydrogen in the compound to determine

the empirical formula.

0.0457 mol H2

1.22 mol compound ≈0.0375

Step 6: Multiply by a factor to obtain whole numbers for the empirical

formula.

0.0375 ×4≈0.15

Step 7: The empirical formula of the compound formed from the reaction is

H2N.

Question 33

Question

A gaseous compound of carbon and hydrogen is burned in excess oxygen gas. If

3.00 L of the compound at STP is burned, what volume of carbon dioxide gas

is produced at the same temperature and pressure? Assume that all gases are

ideal.

Solution

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

pound.

Compound (CxHy) + zO2→pCO2+ qH2O

Step 2: Determine the molar ratio of the reactant and product based on the

balanced equation. Since the compound is burned in excess oxygen, we only

need to consider the reactant compound and the product carbon dioxide. The

coefficients in the balanced equation represent the mole ratio of the reactants

and products. From the balanced equation, 1 mole of the compound produces

1 mole of CO2.

Step 3: Calculate the number of moles of the compound. Given that the

volume of the compound is 3.00 L and the compound is at STP (Standard

Temperature and Pressure), we can use the ideal gas law to find the number of

moles.

P V =nRT

Since the compound is at STP, the pressure is 1 atm, the temperature is 273 K,

and the gas constant R is 0.0821 L·atm/(K·mol):

n=P V

RT =(1.00 atm)(3.00 L)

(0.0821 L ·atm/(K ·mol))(273 K)

Step 4: Calculate the volume of carbon dioxide gas produced. From the

mole ratio determined in Step 2, the number of moles of CO2produced will be

the same as the number of moles of the compound. Use the ideal gas law to

find the volume of carbon dioxide at STP. The pressure and temperature of the

carbon dioxide gas are the same as the original compound, so the volume will

also be 3.00 L.

Therefore, 3.00 L of carbon dioxide gas is produced when 3.00 L of the

compound (CxHy) is burned at STP.

Question 34

Question

A gaseous compound contains only sulfur and fluorine. When 0.348 g of the

compound is reacted with excess hydrogen gas, 71.1 mL of hydrogen fluoride

gas is produced at 295 K and 1.00 atm. What is the empirical formula of the

compound?

(Given: molar volume of a gas at STP = 22.4 L/mol; atomic masses: S =

32.06 g/mol, F = 18.998 g/mol)

Solution

Step 1: Calculate the number of moles of hydrogen fluoride gas produced. Given:

mass of compound = 0.348 g, volume of gas = 71.1 mL, temperature = 295 K,

pressure = 1.00 atm

Convert volume of gas to liters:

Volume of gas (L) = 71.1 mL ×1 L

1000 mL = 0.0711 L

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

produced:

P V =nRT

n=P V

RT =(1.00 atm)(0.0711 L)

(0.0821 L ·atm/mol ·K)(295 K)

n= 0.00244 mol

Step 2: Calculate the number of moles of sulfur in the compound. From the

reaction, we know that 1 mole of the compound reacts with 1 mole of hydrogen

gas to produce 1 mole of hydrogen fluoride gas. Therefore, the number of moles

of sulfur in 0.348 g of the compound is equal to the number of moles of hydrogen

fluoride gas produced.

Thus, moles of sulfur = 0.00244 mol

Step 3: Calculate the atomic ratio of sulfur and fluorine. Let the empiri-

cal formula of the compound be SF. Since the compound contains only sulfur

and fluorine, we have the following equation based on the moles of sulfur and

hydrogen fluoride gas produced:

moles of sulfur

moles of hydrogen fluoride gas =0.00244

0.00244 =1

This indicates that the empirical formula is SF.

Therefore, the empirical formula of the compound is SF.

Question 35

Question

Calculate the volume of nitrogen dioxide gas (NO2) produced when 3.00 moles

of nitric oxide gas (NO) react with excess oxygen gas according to the following

balanced chemical equation:

2NO(g) + O2(g)→2NO2(g)

Solution

Step 1: Write down the balanced chemical equation.

2NO(g) + O2(g)

→2NO2(g)

Step 2: Determine the stoichiometry relating the given and unknown quan-

tities. From the balanced chemical equation, we can see that 2 moles of NO

produce 2 moles of NO2. This means that the ratio of NO to NO2is 1:1.

Step 4: Calculate the mole fractions of oxygen and nitrogen. Since the mole

fraction of oxygen and nitrogen is given by x

x+yand y

x+yrespectively, we get:

x+y=3

x+y=5

Step 5: Substitute the mole fractions into the expressions for partial pres-

sures. Substitute the mole fractions into the expressions for partial pressures to

find:

Poxygen =3

8×1.2 atm

Pnitrogen =5

8×1.2 atm

Step 6: Calculate the partial pressures of oxygen and nitrogen. Finally, solve

for Poxygen and Pnitrogen to find their partial pressures in the gas mixture.

Poxygen = 0.45 atm

Pnitrogen = 0.75 atm

Therefore, the partial pressure of oxygen in the gas mixture is 0.45 atm and

the partial pressure of nitrogen is 0.75 atm.

Question 2

Question

When solid aluminum is placed in hydrochloric acid, the following reaction

occurs:

2Al(s)+6HCl(aq)→2AlCl3(aq)+3H2(g)

If 2.00 grams of aluminum reacts with an excess of hydrochloric acid, what

volume of hydrogen gas is produced at 25

C and 1 atm?

Solution

Step 1: Calculate the number of moles of aluminum used. Given: Mass of

aluminum = 2.00 g, molar mass of aluminum = 26.98 g/mol

Moles of Al =2.00 g

26.98 g/mol ≈0.074mol

Step 2: Determine the limiting reagent. The balanced chemical equation

tells us that 2 moles of aluminum react with 6 moles of HCl to produce 3 moles

of hydrogen gas.

Thus, for 0.074 mol of Al, we would need 6

2×0.074 = 0.222 mol of HCl to

react with it.

In excess HCl, the reaction consumes all the aluminum so HCl is the limiting

reagent.

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

law.

Given: T= 25C= 298 K,Pressure = 1 atm, molar volume of a gas at STP

= 22.4 L/mol

Volume of H2= Moles of H2×Molar volume of gas at STP

= 3 ×0.074 ×22.4≈4.2 L

Therefore, the volume of hydrogen gas produced at 25

C and 1 atm is ap-

proximately 4.2 L.

Question 3

Question

A 2.00 L container is filled with acetylene gas (C2H2) at a pressure of 1.50 atm

and a temperature of 25

C. The acetylene is then burned in oxygen gas to form

carbon dioxide and water vapor according to the following balanced chemical

equation:

2C2H2(g)+5O2(g)→4CO2(g)+2H2O(g)

If the final pressure in the container is 3.00 atm at the same temperature, what

mass of water vapor was produced?

(Note: The molar mass of C2H2is 26.04 g/mol and the molar mass of H2O

is 18.02 g/mol.)

Solution

Step 1: Calculate the moles of acetylene gas using the ideal gas law.

PV = nRT

n = PV

RT =(1.50 atm)(2.00 L)

0.0821 L ·atm/mol ·K·(25 + 273) K

n≈0.124 mol

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

tion, we can see that 2 moles of acetylene produce 2 moles of water. Since each

mole of acetylene produces 2 moles of water, only 0.062 moles of water will be

produced.

Step 3: Calculate the volume of water vapor at the final pressure using the

ideal gas law.

n = PV

V = nRT

P=(0.062 mol)(0.0821 L ·atm/mol ·K·(25 + 273) K)

3.00 atm

V≈0.341 L

Step 4: Convert the volume of water vapor to mass using the molar mass of

water.

Mass = n ×molar mass = (0.062 mol)(18.02 g/mol)

Mass = 1.12 g

Therefore, approximately 1.12 grams of water vapor were produced during

the reaction.

Question 4

Question

A reaction between solid lithium hydroxide and gaseous carbon dioxide produces

solid lithium carbonate and water vapor. If 5.00 g of lithium hydroxide reacts

with excess carbon dioxide to produce 2.67 g of water vapor, what is the limiting

reactant? What is the theoretical yield of lithium carbonate in grams?

Solution

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

hydroxide (LiOH) and carbon dioxide (CO2).

LiOH(s) + CO2(g)→Li2CO3(s)+H2O(g)

Step 2: Calculate the molar mass of lithium hydroxide (LiOH).

Molar mass of LiOH = Molar mass of Li + Molar mass of O + Molar mass of H

Molar mass of LiOH = 6.94 g/mol + 15.999 g/mol + 1.008 g/mol = 23.948 g/mol

Step 3: Calculate the moles of water vapor produced.

Moles of H2O = Mass of H2O

Molar mass of H2O=2.67 g

18.015 g/mol

Moles of H2O≈0.148 mol

Step 4: Identify the stoichiometry of the reaction from the chemical equation.

From the balanced chemical equation, we see that 1 mole of lithium hydroxide

reacts with 1 mole of carbon dioxide to produce 1 mole of water vapor.

Step 5: Use the stoichiometry of the reaction to calculate the moles of lithium

hydroxide reacted. Since 1 mole of lithium hydroxide produces 1 mole of water

vapor, the moles of lithium hydroxide reacted are also approximately 0.148 mol.

Step 6: Calculate the theoretical yield of lithium carbonate. From the bal-

anced chemical equation, the molar ratio between lithium hydroxide and lithium

carbonate is 2:1.

Moles of Li2CO3= 0.148 mol ×1 mol Li2CO3

2 mol LiOH = 0.074 mol

Theoretical yield of Li2CO3= 0.074 mol ×Molar mass of Li2CO3

Step 7: Calculate the theoretical yield of lithium carbonate in grams.

Theoretical yield of Li2CO3= 0.074 mol ×73.89 g/mol = 5.47 g

Step 8: Conclusion The limiting reactant in the reaction is lithium hydroxide,

and the theoretical yield of lithium carbonate is 5.47 grams.

Question 5

Question

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

dioxide and water vapor. If 2.00 L of the hydrocarbon at STP (standard tem-

perature and pressure) produces 9.00 L of carbon dioxide, what is the molecular

formula of the hydrocarbon?

Solution

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

CxHy+ (x + y/4)O2−−→ xCO2+ (y/2)H2O

Step 2: Calculate the number of moles of carbon dioxide produced. Given:

Volume of hydrocarbon = 2.00 L Volume of carbon dioxide produced = 9.00 L

Since we are at STP (standard temperature and pressure), we can use the

volume ratios as mole ratios. Moles of carbon dioxide = Volume of carbon

dioxide / 22.4 L Moles of carbon dioxide = 9.00 L / 22.4 L = 0.40179 mol

Step 3: Determine the mole ratio of carbon dioxide to the hydrocarbon.

From the balanced equation, the mole ratio between carbon dioxide and the

hydrocarbon is 1:1.

Step 4: Determine the number of moles of the hydrocarbon. Since the mole

ratio between carbon dioxide and the hydrocarbon is 1:1, the number of moles

of the hydrocarbon is the same as the number of moles of carbon dioxide. Moles

of hydrocarbon = 0.40179 mol