WS 7: Gas Stoichiometry Problems
1. Solid iron(III) hydroxide decomposes to produce iron(III) oxide and water vapor.
a. Write the balanced equation: 2Fe(OH)3  Fe2O3 + 3H2O
15.0 g
xL
b. If 15.0 grams of iron(III) hydroxide completely decompose, how many liters of water vapor are
produced at STP?
15.0g Fe(OH)3 ×
1 mol Fe(OH)3
3 mol H2 O
22.4 L H2 O
×
×
= 4.72 L H2 O
106.88 g Fe(OH)3
2 mol Fe(OH)3
1 mol H2 O
2. A chemist reacts 25.0 grams of solid iron with excess sulfuric acid to produce iron(II) sulfate and
hydrogen gas.
a. Write the balanced equation: Fe(s) + H2SO4  FeSO4 + H2
25.0 g
xL
b. What will be the volume of hydrogen gas collected at 1.50 atm and 298 K?
25.0g Fe ×
𝑉=
1 mol Fe
1 mol H2
×
= 0.448 mol H2
55.85 g Fe
1 mol Fe
𝑛𝑅𝑇
(0.448 mol)(0.08206)(298 K)
=
= 7.30 𝐿 H2
𝑃
1.50 atm
Problems 3 and 4 are repeats of above
5. Propane gas (C3H8) burns in the presence of oxygen gas to produce carbon dioxide and water
vapor. The reaction takes place at 0.950 atm and 325 K.
a. Write the balanced equation. C3H8 + 5O2  3CO2 + 4H2O
3.25 g
xL
xL
xL
b. If 3.25 grams of propane is burned, how many liters of CO2 are produced?
3.25g C3 H8 ×
𝑉=
1 mol C3 H8
3 mol CO2
×
= 0.221 mol CO2
44.09g C3 H8
1 mol C3 H8
𝑛𝑅𝑇
(0.221 mol)(0.08206)(325 K)
=
= 6.20 𝐿 CO2
𝑃
0.950 atm
c. What volume of H2O vapor is produced?
3.25g C3 H8 ×
𝑉=
1 mol C3 H8
4 mol H2 O
×
= 0.295 mol H2 O
44.09g C3 H8
1 mol C3 H8
𝑛𝑅𝑇
(0.295 mol)(0.08206)(325 K)
=
= 8.28 𝐿 H2 O
𝑃
0.950 atm
d. What volume of O2 is required?
3.25g C3 H8 ×
𝑉=
1 mol C3 H8
5 mol O2
×
= 0.369 mol O2
44.09g C3 H8
1 mol C3 H8
𝑛𝑅𝑇
(0.369 mol)(0.08206)(325 K)
=
= 10.4 𝐿 O2
𝑃
0.950 atm
6. A 3.25-gram sample of solid calcium carbide (CaC2) reacts with water to produce C2H2 gas and
aqueous calcium hydroxide.
a. Write the balanced equation. CaC2 + 2H2O  C2H2(g) + Ca(OH)2
3.25 g
xL
b. How many liters of C2H2 product are collected at 293 K and 80.00 kPa? 80.00 kPa = 0.790 atm
3.25g CaC2 ×
𝑉=
1 mol CaC2
1 mol C2 H2
×
= 0.0507 mol C2 H4
64.10g CaC2
1 mol CaC2
𝑛𝑅𝑇
(0.0507 mol)(0.08206)(293 K)
=
= 1.54 𝐿 C2 H2
𝑃
0.790 atm
7. Ammonium sulfate, an important fertilizer, can be prepared by the reaction of ammonia with
sulfuric acid according to the following balanced equation:
2 NH3(g) + H2SO4(aq)  (NH4)2SO4(aq)
Calculate the volume of NH3 (in liters) needed to react with 150.0 kg of H2SO4 if the reaction occurs at
20.0ºC and 25.0 atm.
150.0 kg H2 SO4 ×
𝑉=
1000g
1 mol H2 SO4
2 mol NH3
×
×
= 3058 mol NH3
1kg
98.09g H2 SO4
1 mol H2 SO4
𝑛𝑅𝑇
(3058 mol)(0.08206)(293 K)
=
= 2940 𝐿 NH3
𝑃
25.0 atm
8. Assume that 8.50 L of chlorine gas are consumed at 298 K and 243,180 Pa according to the
following balanced equation:
2KI(aq) + Cl2(g)  2KCl(aq) + I2(s)
8.50 L
xg
How many grams of I2 are produced?
243,180 Pa = 2.40 atm
You have to convert L of Cl2 into moles of Cl2 first. Then solve as a moles  moles  grams problem.
𝑛=
𝑃𝑉
(2.40 atm)(8.50 L)
=
= 0.834 moles Cl2
𝑅𝑇
(0.80206)(298 K)
0.834 mol Cl2 ×
1 mol I2
253.8 g I2
×
= 212 g I2
1 mol Cl2
1 mol I2