Theoretical Yield Problem?

Question

Isoctane Balanced Equation: C8H18 + 11O2 --> 8CO2 + 6H2O

Assuming gasoline is 100% isoctane with density 0.692 g/mL. What is the theoretical yield of carbon dioxide produced by the combustion of 1.2 * 10^10 gal gasoline (approx. annual consumption of gasoline in U.S)?

Answer 1

C8H18 ==> 8*12 + 18 = 114 amu
8CO2 ==> 8*12+16*16 = 352 amu
1 US gallon = 3,785.4118 ml
(1.2*10^10 gal)(3,785.4118 ml/gal)(0.692g/ml)(352/114) = 97,059,552,409,600 g ≈
9.7*10^7 metric tons
The mass of Earth's atmosphere is about
5.1*10^18 kg or 5.1*10^15 metric tons
(100)(9.7*10^7)/(5.1*10^15) ≈ 1.9*10^-6 % (0.0000019%)
100(0.0000019% / 0.0384%) = 0.0050% increase in atmospheric CO2

Answer 2

this may well be a limiting reactant undertaking you are able to desire to verify how a lot SO3 may well be created from 2SO2 and O2 use ingredient prognosis for SO2... commence with .292 L SO2 X (2 mol SO3 / 2 mol 2SO2) = .292 L SO3 for O2... .1542 L O2 X ( 2 mol SO3 / a million mol O2) = .3084 L SO3 SO2 is the limiting reactant b/c there is completely adequate SO2 to make .292 L of SO3 so the theoretical yield could be 292 ml