Question

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)?

Answer 1

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
`1.453* 10^(14)g`

Explanation : Given,

Volume of isooctane =
`1.80* 10^(10)gallons`

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

1 gallon = 3.785 L

So,
`1.80* 10^(10)gallon* ((3.785L)/(1gallon))=6.813* 10^(10)L`

Now we have to calculate the mass of isooctane.

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Volume of isooctane =
`6.813* 10^(10)L=6.813* 10^(13)mL` (Conversion factor: 1 L = 1000 mL)

Density of isooctane = 0.692 g/mL

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

`0.692g/mL=\frac{\text{Mass of isooctane}}{6.813* 10^(13)mL}\\\\\text{Mass of isooctane}=(0.692g/mL* 6.813* 10^(13)mL)=4.714* 10^(13)g`

Now we have to calculate the number of moles.

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

Given mass of isooctane =
`4.714* 10^(13)g`

Molar mass of isooctane = 114.22 g/mol

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

`\text{Moles of isooctane}=(4.714* 10^(13)g)/(114.22g/mol)=4.127* 10^(11)mol`

The chemical equation for the combustion of isooctane is:

`2C_8H_(18)+25O_2\rightarrow 16CO_2+18H_2O`

By Stoichiometry of the reaction we can say that,

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

So,
`4.127* 10^(11)mol` of isooctane will produce =
`(16)/(2)* 4.127* 10^(11)mol=3.3016* 10^(12)mol` 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 =
`3.3016* 10^(12)mol`

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

`3.3016* 10^(12)mol=\frac{\text{Mass of carbon dioxide}}{44.00g/mol}\\\\\text{Mass of carbon dioxide}=(3.3016* 10^(12)mol* 44.00g/mol)=1.453* 10^(14)g`

Hence, the theoretical yield of carbon dioxide is
`1.453* 10^(14)g`