gasoline combustion chemistry question?

Question

I would appreciate someone explaining the steps needed to solve this so I can do it myself. I don't just want an answer. Thanks.

A clean burning engine will emit about 5.5 lbs of carbon in the form of CO2 for every gallon of gas it consumes. Illistrate this statement to be mathmatically correct by using the following information:
a) one of the main components of gasoline is octane (C8H18) with a density of 0.692 g/mL. 1.00 gallon has the same volume as 3,790.00 ml. 454.00 grams = 1.00 lb.
b) another component of gasoline is toluene (C7H8) with a density of 0.867 g/mL.

Answer 1

When 1 mol of octane is burned the following equation obtains.
C8H18 + 12.5 O2 = 8 CO2 + 9 H2O

Thus for every mol of octane burned 8 mols of CO2 will form and therefore eight mols of C will be in the air

1.0 gal x 3790 ml/gal x.692 g/ml x = 2616g of octane.

Mols of octane = 2616g/114 = 22.9 mols octane
mols CO2 = 8 x 22.9 = 183.2 mol CO2

Mass CO2 = 183.2 x 44 g/mol x1lb/454g =17.7 lbs CO2

Mass C = 17.7 lbs (12/44) = 4.82g C

Fro par B you will need the formula for toluene and it is C7H8 Therefore for each mol of toluene that is burned you will form 7 mol ovf co2 and 7 mols of C

Do the same calculation as above and you will get the answer.
Mass C = 177,2 g CO2/ 44(g/mol) =

Answer 2

Find both mols of gasoline and toluene.

By one, or two balanced combustion equations do the stoichiometry to see how much CO2 produced. Convert into pounds.

Answer 3

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

Atomic weights: C=12 O=16 H=1 CO2=44 C8H18=114

Let gasoline be called G

1galG x 3790mLG/1galG x 0.692gG/1mLG x 1molG/114gG x 16molCO2/2molG x 44gCO2/1molCO2 x 1lbCO2/454gCO2 = 17.8 lb CO2