Gas Stoichiometry

QuickNotes

Gas Stoichiometry

• Gas stoichiometry involves using balanced chemical equations and given/known values of various elements of the ideal gas law to obtain a needed value. This can primarily be shown through examples.
• Important Values:
• At STP:
  P = 1 atm
  V = 22.4 L
  T = 273 K

Examples: 

Example 1: A sample of nitrogen gas has a volume of 2.25 L at STP. How many moles of nitrogen gas are present?

• Recognize that 1 mole of nitrogen gas at STP has a volume of 22.4 L.

PV=nRT

(1 atm)(x L) = (1 mol)(0.08206 L atm/K mol)(273 K)

x = 22.4 L

• Use the given volume of gas (2.25 L) and 1 mol N2 = 22.4 L to find the moles of gas:

(2.25 L) x (1 mol/22.4 L) = 0.100 mol N2 

Example 2: 2.35 L of oxygen gas reacts with 3.72 L of hydrogen gas, forming water. How many liters of the excess reactant will remain? If 2.50 L of water were actually produced, what would be the percent yield?

• The balanced chemical equation is:

2H₂ + O₂ ---> 2H₂O

• Determine the limiting reagent:

oxygen: 2.35 / 1 = 2.35

hydrogen: 3.72 / 2 = 1.86

Hydrogen is the limiting reagent.

Please note that no mention of temperature or pressure is made in the problem. This means that everything takes place at an unchanged temperature and pressure. Consequently, those two values remain constant and drop out of consideration. We do not need moles because, in a situation of constant temperature and pressure, the volumes are directly proportional to the number of moles.

Besides which, we cannot even calculate moles since we do not know the temperature or the pressure.

• Use H₂:O₂ molar ratio:

2 is to 1 as 3.72 is to x, where x = 1.86 L of oxygen used.

2.35 - 1.86 = 0.49 L of unreacted oxygen remaining

•  Percent yield:

3.72 L of water are produced.2.50 / 3.72 = 67.2%

Example 3: How much air is needed (in m³, at 25.0 °C, 1.00 atm) to completely burn 10.0 moles of propane (C₃H₈)? Assume that the air is composed of 21.0% O_2.

• The combustion of propane:

C₃H₈ + 5O₂ ---> 3CO₂ + 4H₂O

• Determine moles of pure oxygen needed to burn 10.0 mol of propane:

1 is to 5 as 10.0 is to x, where x = 50.0 mol of oxygen required

• Use PV = nRT to convert mol of oxygen to liters of oxygen at the conditions stated in the problem:

(1.00 atm) (V) = (50.0 mol) (0.08206 L atm mol¯¹ K¯¹)(298 K)

V = 1222.694 L

• Convert to volume of air required:

1222.694 L / 0.21 = 5822.35 L

• Convert to cubic meters:

5822.35 L = 5822.35 dm³

1 m³ = 1000 dm³

Therefore, the answer is 5.82 m³

Example 4: Nitrogen monoxide reacts with oxygen according to the equation below:

2NO(g) + O_­2(g) --> NO_­2(g)

How many liters of NO (reacting with excess oxygen) are required to produce 3.0 liters of NO_­2?

• State ratio of volumes. In this case, it is 2:1
• Write a ratio and proportion:

2 is to 1 as x is to 3.0 L

x = 6.0 L of NO required 

Be sure and review this video to fully understand gas stoichiometry! Watching a run-through of these problems can clarify how to begin their setups.  

Check out this mini quiz to make sure you can do gas stoichiometry!

Gas Stoichiometry Quiz