Stoichiometry and Diffusion Problems
1. Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.
Rate of effusion H2 = √MO2 = √32 = √16 = 4
Rate of effusion O2 √MH2
√2
H2 diffuses 4 times faster than O2
2. A sample of hydrogen gas effuses through a porous container 8.94 times faster than an unknown gas. What is the
molar mass of the unknown gas?
Rate of effusion H2
= √Munkn
Rate of effusion Unkn √MH2
8.94x = √Munkn
x
√2
Munkn = [(8.94)(√2)]2
Munkn = 160 g/mol
(√Munkn)=(8.94)(√2)
3. Calculate the molar mass of a gas that effuses at 1.59 times the effusion rate of carbon dioxide.
Rate of effusion CO2 = √Munkn
Rate of effusion Unkn
√MCO2
x = √Munkn
1.59x √44
(√44) = (1.59)(√Munkn)
(√Munkn) = (√44)/(1.59)
Munkn = [(√44)/(1.59)]2
Munkn = 17.4 g/mol
4. The complete combustion or propane (C3H8), a gas that is in tanks and used for cooking and heating in some
areas, occurs according to the following equation:
C3H8 (g) + 5O2 (g)  3 CO2 (g) + 4H2O (g)
Assuming all volume measurements are made at the same temperature and pressure, (a) What will be the volume in
liters of oxygen required for the complete combustion of 0.350 L of propane; and b) What will be the volume of carbon
dioxide produced in the reaction? This is at STP!! “same temperature and pressure = at STP” Therefore, you can use
the 1 mol = 22.4 L conversion.
a) 0.350 L C3H8 x 1 mol C3H8 x 5 mol O2 x 22.4 L  1.75 L O2
22.4 L
1 mol C3H8
1 mol O2
b) 0.350 L C3H8 x 1 mol C3H8 x 3 mol CO2 x 22.4 L  1.05 L CO2
22.4 L
1 mol C3H8
1 mol CO2
5. Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen
gas is needed to react completely with 4.55 L of oxygen gas to produce water vapor? 2 H2 + O2  2 H2O
This is at STP!! “same temperature and pressure = at STP” Therefore, you can use the 1 mol = 22.4 L conversion.
4.55 L O2 x
1 mol O2 x
22.4 L O2
2 mol H2
1 mol O2
x
22.4 L  9.1 L H2
1 mol H2
6. Calcium carbonate (CaCO3), which is also known as limestone, is calcined to produce calcium oxide (lime), an
industrial chemical with a wide variety of uses ranging from glass manufacture to sewage treatment. The
balanced equation is: CaCO3 (s)  CaO (s) + CO2 (g)
How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP?
5 L CO2 x
1 mol CO2 x
22.4 L CO2
1 mol CaCO3
1 mol CO2
x
100g CaCO3  22.3 g CaCO3
1 mol CaCO3
7. Tungsten, a metal that is used in light bulb filaments, is produced industrially by the following reaction of
tungsten oxide with hydrogen: WO3 (s) + 3H2 (g)  W (s) + 2 H2O (l)
How many liters of hydrogen at 35°C and 745 mm Hg are needed to react completely with 875 g tungsten oxide?
***This is NOT at STP therefore you have to use the Ideal Gas Law to solve for Volume. PV = nRT You are
given pressure, temperature, and you know R, but you need the number of moles (n) in order to solve for
volume. Therefore you convert grams of tungsten oxide (what you are starting with) into moles of hydrogen
(since you are solving for a volume of hydrogen, you need to have your moles in hydrogen).
875 g WO3 x
1 mol WO3 x
232 g WO3
3 mol H2  11.3mol H2 = n
1 mol WO3
Also remember when using the Ideal Gas Law, pressure must be in atm, temperature must be in Kelvin, and
Volume is in Liters. R is always equal to 0.0821.
745 mm Hg x
1 atm
= 0.98 atm
760 mm Hg
35° = 308 K
PV = nRT
(0.98)(V) = (11.3)(0.0821)(308)
V = 292 L H
8. What mass of sulfur must be used to promote 12.61 L of sulfur dioxide at STP according to the equation.
S (s) + O2 (g)  SO2
12.61 L SO2 x
1 mol SO2 x
22.4 L SO2
1 mol S x
1 mol SO2
34g S  18.0 g S
1 mol S
9. How many liters of carbon monoxide at 27°C and 188 mm Hg can be produced from the burning of 65.5 g of
carbon according to the equation: 2 C (s) + O2 (g)  2 CO (g)
***This is NOT at STP therefore you have to use the Ideal Gas Law to solve for Volume. PV = nRT You are
given pressure, temperature, and you know R, but you need the number of moles (n) in order to solve for
volume. Therefore you convert grams of carbon (what you are starting with) into moles of carbon monoxide
(since you are solving for a volume of carbon monoxide, you need to have your moles in carbon monoxide).
65.5 g C x
1 mol C x
12 g C
2 mol CO
2 mol C
 5.46 mol CO = n
Also remember when using the Ideal Gas Law, pressure must be in atm, temperature must be in Kelvin, and
Volume is in Liters. R is always equal to 0.0821.
188 mm Hg x
1 atm
= 0.247 atm
760 mm Hg
PV = nRT
(0.247)(V) = (5.46)(0.0821)(300)
V = 544 L CO
27° = 300 K
10. How many liters of ammonia gas can be formed from the reaction of 150.0 L of hydrogen gas, assuming that
there is complete reaction of hydrogen with excess nitrogen gas and that all measurements are made at the same
temperature and pressure?
This is at STP!! “same temperature and pressure = at STP” Therefore, you can use the 1 mol = 22.4 L conversion.
Also, on the test I will give you the equations, so you will not have to generate them yourself.
N2 + 3 H2  2 NH3
150 L H2 x
1 mol H2 x
22.4 L H2
x 22.4 L  100 L NH3
1 mol NH3
2 mol NH3
3 mol H2
11. How many liters of H2 at STP can be produced by the reaction of 4.60 g of Na and excess water?
This is at STP!! Therefore, you can use the 1 mol = 22.4 L conversion.
Also, on the test I will give you the equations, so you will not have to generate them yourself.
2Na + H2O  Na2O + H2
4.60 g Na x 1 mol Na x
23 g Na
1 mol H2 x 22.4 L  2.24 L H2
2 mol Na
1 mol H2
12. How many grams of Na are needed to react with H2O to liberate 400 mL of H2 at STP?
This is at STP!! Therefore, you can use the 1 mol = 22.4 L conversion.
Also, on the test I will give you the equations, so you will not have to generate them yourself.
2Na + H2O  Na2O + H2
Note: 400 mL x
0.400 L H2 x
1L
= 0.4 L
1000 mL
1 mol H2 x
22.4 L H2
2 mol Na
1 mol H2
x 23 g Na  0.821 g Na
1 mol Na
13. What volume of oxygen in liters can be collected at 750 mm Hg pressure and 25°C when 30.6 g of KClO3
decomposes by heating?
Here is the equation (on the test, I will give you the equation): 2 KClO3  2 KCl + 3O2
***This is NOT at STP therefore you have to use the Ideal Gas Law to solve for Volume. PV = nRT You are
given pressure, temperature, and you know R, but you need the number of moles (n) in order to solve for
volume. Therefore you convert grams of KClO3 (what you are starting with) into moles of oxygen (since you are
solving for a volume of oxygen, you need to have your moles in oxygen).
30.6 g KClO3 x
1 mol KClO3
123 g KClO3
x
3 mol O2  0.373 mol O2 = n
1 mol KClO3
Also remember when using the Ideal Gas Law, pressure must be in atm, temperature must be in Kelvin, and
Volume is in Liters. R is always equal to 0.0821.
750 mm Hg x
1 atm
= 0.987 atm
760 mm Hg
PV = nRT
(0.987)(V) = (0.373)(0.0821)(298)
V = 9.25 L O2
25° = 298 K
14. When a 115 gram piece of aluminum metal is added to excess sulfuric acid, how many liters of hydrogen gas are
formed at STP? 2 Al + 3 H2SO4  Al2(SO4)3 + 3 H2
115 g Al x
1 mol Al x
27 g Al
3 mol H2
2 mol Al
x
22.4 L  143 L H2
1 mol H2
15. If 2 L of hydrogen gas were formed at 1.25 atm and 27°C, what was the mass of aluminum that was used? 2 Al +
3 H2SO4  Al2(SO4)3 + 3 H2
You are looking for mass in this case. You have volume, pressure, and temperature given to you, so you can solve for the
number of moles (n) using the Ideal Gas Law. From there, you can convert moles to grams.
27°C = 300 K
PV = nRT
(1.25)(2) = (n)(0.0821)(300)
n = 0.102 mol H2
0.102 mol H2 x
2 mol Al x
3 mol H2
27 g Al  1.84 g Al
1 mol Al