Theoretical yield of CO2 produced by combustion

Question

Assuming gasoline is 90.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO₂ produced by the combustion of 1.10 x 10¹⁰ gallons of gasoline (the estimated annual consumption of gasoline in the U.S.)? Remember, there are 3.785 liters in 1 gallon and assume that isooctane is the only carbon containing component of gasoline.

Scientific notation can be entered as follows: 1.23 x 10²³ = 1.23E23

The answer I came up with was 1.59E14, but it was wrong.

Answer 1

Iso-octane = C₈H₁₈

2C₈H₁₈ + 25O₂ ==> 16CO₂ + 18H₂O ... balanced equation

First, let's find liters of iso-octane, and then convert that to grams using the density:

liters C₈H₁₈ = 1.10x10¹⁰ gal x 3.785 L/gal = 4.164x10¹⁰ L

grams C₈H₁₈ = 4.164x10¹⁰ L x 692 g/L = 2.881x10¹³ g

Now we can correct for the 90.0% and then convert to moles:

2.881x10¹³ g x 90% = 2.5929x10¹³ g

moles C₈H₁₈ = 2.5929x10¹³ g x 1 mol/114.22 g = 2.2745x10¹¹ moles

Theoretical yield of CO₂ = 2.2745x10¹¹ moles C₈H₁₈ x 16 mole CO₂/2 mol C₈H₁₈ x 44.01 g/mol = 8.00x10¹¹ g