stoichiometry (chemistry) problem?

Question

a typical car gets 30 mpg of gas & drives 12000 mi per yr. assuming that octane (C8H18 density 0.07025) is a principal component of gasoline;
a. how much oxygen is required for a car to run for 1 yr?
b. how many grams of CO2 is produced by a car in 1yr?

the chemical formula for gasoline is :
2 C8H18 + 25 O2 -> 18 H2O + 16 CO2

Answer 1

OK, we start with gallons and convert over to grams. Note your density is low (should be 0.7025). We next go to moles. The equation is used to find the oxygen and the CO2. I would imagine that the oxygen required is also in grams. It's dogwork, but pretty straightforward.

12000 miles/yr / 30 miles/gallon = 400 gallon per year.
The gallon is the volume that contains 7.48 pounds of H2O, or 3394 grams. With the given density, each gallon would contain 3394x.7025 or 2384 grams of octane.

Now to the reaction. The mol weight of octane is 114 or
8x12 + 18x1. So each gallon contains (2384/114) moles of octane or 20.91 moles. The reaction says that 25 moles of oxygen (as the dimer) goes with 2 moles of octane.
So the oxygen moles required to combust a gallon of octane is (25/2)x20.91 =261 moles. Each mole weights 32 g, so the oxygen requirement per gallon is 8364 grams. For a years worth of gas, or 400 gallons, we have 3.35x10^6 grams of oxygen.

Back to our chemical equation again, and for each 25 moles of oxygen used, 16 moles of CO2 are produced. From above, a gallon of octane uses 261 moles of O2. So moles of CO2/gallon of octane = 261 x 16/25 = 167 moles. Each mole weights 44 grams, so the weight is 7350 grams of CO2, and for 400 gallons, the weight is
2.93x10^6 g/year.

Again, the key to this (and similar problems) is to get mass converted to moles and to use the reaction to determine the moles of other constituents of the reaction, and then convert back.

Answer 2

12000mi/30mpgx4.546(L/gallon)
x70.25(g/L)/(8x12+18)(g/mol)
x(25/2)x32(g/mol)x10E-6
=0.448 t