Answer :
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 [tex]1.453\times 10^{14}g[/tex]
Explanation : Given,
Volume of isooctane = [tex]1.80\times 10^{10}gallons[/tex]
First we have to convert volume into liters, we use the conversion factor:
1 gallon = 3.785 L
So, [tex]1.80\times 10^{10}gallon\times (\frac{3.785L}{1gallon})=6.813\times 10^{10}L[/tex]
Now we have to calculate the mass of isooctane.
[tex]\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}[/tex]
Volume of isooctane = [tex]6.813\times 10^{10}L=6.813\times 10^{13}mL[/tex] (Conversion factor: 1 L = 1000 mL)
Density of isooctane = 0.692 g/mL
Now put all the given values in above equation, we get:
[tex]0.692g/mL=\frac{\text{Mass of isooctane}}{6.813\times 10^{13}mL}\\\\\text{Mass of isooctane}=(0.692g/mL\times 6.813\times 10^{13}mL)=4.714\times 10^{13}g[/tex]
Now we have to calculate the number of moles.
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] .....(1)
Given mass of isooctane = [tex]4.714\times 10^{13}g[/tex]
Molar mass of isooctane = 114.22 g/mol
Now put all the given values in equation 1, we get:
[tex]\text{Moles of isooctane}=\frac{4.714\times 10^{13}g}{114.22g/mol}=4.127\times 10^{11}mol[/tex]
The chemical equation for the combustion of isooctane is:
[tex]2C_8H_{18}+25O_2\rightarrow 16CO_2+18H_2O[/tex]
By Stoichiometry of the reaction we can say that,
As, 2 moles of isooctane produces 16 moles of carbon dioxide.
So, [tex]4.127\times 10^{11}mol[/tex] of isooctane will produce = [tex]\frac{16}{2}\times 4.127\times 10^{11}mol=3.3016\times 10^{12}mol[/tex] 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 = [tex]3.3016\times 10^{12}mol[/tex]
Now put all the given values in equation 1, we get:
[tex]3.3016\times 10^{12}mol=\frac{\text{Mass of carbon dioxide}}{44.00g/mol}\\\\\text{Mass of carbon dioxide}=(3.3016\times 10^{12}mol\times 44.00g/mol)=1.453\times 10^{14}g[/tex]
Hence, the theoretical yield of carbon dioxide is [tex]1.453\times 10^{14}g[/tex]