Can some1 help with a difficult Stoichiometry question?

Question

My car has a 13.7 gallon tank. Assuming that gasoline is entirely made from octane(C8H18), how many POUNDS of CO2 would be froduced from my car each time I use a tank of gas? How many pounds of CO2 will produce from my car each year(I drive 21,500 miles each year)?
Some equalities you may find useful
- 1 lbs- 4536 kg
1 kg-2.205 lbs.
28.35g-3.785 liters
l inch-2.54 cm
density of octane-.855g/ml
my car averages 29.7 miles/gallon

use the following equation
C8H18 + 12 1/2 O2 --> 8 CO2 + 9H2O
if u can find the answer and show work, it would be greatly appreciated. Im a chem teacher and a student gave this prob. to me. If i get stumped by my student, they may treat me w/ less respect. Please hlp. thx

the mileage is given on the bottom

Steve, the milege is there

Answer 1

C8H18, MW = 114. CO2, MW = 44. Molecular weights are relative, so you can have 114g/1 g-mol or 114lb in 1 lb-mol. Let gasoline be called G

13.7galG x 5.9lbG/1galG x 1lb-molG/114lbG x 8lb-molCO2/1lb-molG x 44lbCO2/1lb-molCO2) = (13.7)(5.9)(8)(44)/(114) = 250lb CO2

The 13.7galG is given. The first factor is the density of octane in lb/gal. The galG cancel, leaving lbG.

For the future, you may want to memorize 8.320lb/gal, which is the density of water in English units. If you know the specific gravity of any liquid, which happens to equal the density in g/mL, then multiplying it by the density of water gives you the density of the other liquid in lb/gal.

The second factor comes from the molecular weight of octane. The lbG cancel, leaving lb-molG. The third factor comes from the balanced equation. The lb-molG cancel, leaving lb-molCO2. The fourth factor comes from the molecular weight of CO2. The lb-molCO2 cancel, leaving lbCO2.

For the next part of the question, you need to invoke more factors involving the mileage you get, etc.

Answer 2

Where do you summise that 28.35 g = a gallon?

All this is, is a factor label method problem. Don't you guys teach that around the first week of class or so? Makes me wonder if you're a teacher.... or a student masquerading as one.
As most chemistry texts go over this and furiously.

Nice try, my friend.

Answer 3

Actually not all of this info are necessary. 13.7 gallon is approx 51.861 litres. mutliplying by density of octane the quantity of octane is 44.341 kg or 388.9 moles. So it reacts to produces 8*388.9 = 3111 moles of carbon di oxide. Which is 136.91 kg.