Assuming gasoline is isooctane, with a density of g/mL, what is the theoretical yield of carbon dioxide produced by the combustion of gal of gasoline (the approximate annual consumption of gasoline in the United States)?

This is an incomplete question, here is a complete question.

Assuming gasoline is 89.0% isooctane, with a density of 0.692 g/mL, what is the theoretical yield (in grams) of CO₂ produced by the combustion of 1.80 × 10¹⁰ gallons of gasoline (the estimated annual consumption of gasoline in the U.S.)?

Answer : The theoretical yield of carbon dioxide is

Explanation :  Given,

Volume of isooctane =

First we have to convert volume into liters, we use the conversion factor:

1 gallon = 3.785 L

So,

Now we have to calculate the mass of isooctane.

Volume of isooctane =    (Conversion factor:  1 L = 1000 mL)

Density of isooctane = 0.692 g/mL

Now put all the given values in above equation, we get:

Now we have to calculate the number of moles.

     .....(1)

Given mass of isooctane =

Molar mass of isooctane = 114.22 g/mol

Now put all the given values in equation 1, we get:

The chemical equation for the combustion of isooctane is:

By Stoichiometry of the reaction we can say that,

As, 2 moles of isooctane produces 16 moles of carbon dioxide.

So, of isooctane will produce = of carbon dioxide

Now we have to calculate the mass of carbon dioxide.

Molar mass of carbon dioxide = 44.00 g/mol

Moles of carbon dioxide =

Now put all the given values in equation 1, we get:

Hence, the theoretical yield of carbon dioxide is

50.23 i believe is correct but don't quote me on that.