Panel 13: A gas stoichiometry problem

An automobile is driven by chemical reactions which can be reasonably modeled by the combustion of octane. It takes about one mole of octane to move a car one mile.

What volume of oxygen at 300K and 1.00 atm is required to react with one mole of octane according to the reaction given below?

2 C8H18 + 25 O2 -> 16 CO2 + 18 H2O

What volume of air is required, given that air is 21% oxygen?

Solution step 1: From the reaction stoichiometry, 25/2 moles of oxygen are required to burn one mole of octane.

Solution step 2: Determine the volume of 25/2 moles of oxygen at 300K and 1.00 atm.
V = nRT/P = ( (12.5 mol)(0.0821L atm mol-1 K-1)(300. K) ) /(1.00 atm) = 308 L

Solution step 3: Determine the volume of air required.
Vair = (308L O2) x (100L air)/(21L O2) = 1470 L air