What volume of air at 1.0 atm and 25 degree Celsius must be taken to an engine to burn 1 gallon of gasoline?

Question

Assuming that:

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

provides reasonable model of the actual combustion process. ( The partial pressure of oxygen in air is 0.21 atm and the density of liquid octane is 0.70g/mL)

Answer 1

1 gallon = 3.785 L octane.
mass octane = volume x density = approx 0.7 x 3785 mL = approx 250 g
mols octane = grams/molar mass = approx 23

23 mols C8H18 x (25 mols O2/2 mol octane) = approx 300 mols oxygen needed. Check all of these numbers.
Then use PV = nRT to solve for volume oxygen needed and correct for the 21% O2 in air to find volume air needed.

Answer 2

thanks DrBob222