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 x 1010 gallons of gasoline (the estimated annual consumption of gasoline in the u. remember, there are 3.785 liters in 1 gallon and assume that isooctane is the only carbon containing component of gasoline. scientific notation can be entered as follows: 1.23 x 1023 = 1.23e23

The theoretical yield of carbon dioxide is

Explanation:

We are given:

Volume of isooctane =

To convert this into liters, we use the conversion factor:

1 gallon = 3.785 L

So,

To calculate the mass of isooctane, we use the equation:

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

Density of isooctane = 0.692 g/mL

Putting values in above equation, we get:

To calculate the number of moles, we use the equation:

     .....(1)

Given mass of isooctane =

Molar mass of isooctane = 114.22 g/mol

Putting values in equation 1, we get:

The chemical equation for the combustion of isooctane follows:

By Stoichiometry of the reaction;

2 moles of isooctane produces 16 moles of carbon dioxide.

So, of isooctane will produce = of carbon dioxide

Molar mass of carbon dioxide = 44.00 g/mol

Moles of carbon dioxide =

Putting values in equation 1, we get:

Hence, the theoretical yield of carbon dioxide is

i'm pretty sure its b i might be wrong though!