a. Write the balanced equation for the combustion of isooctane \(\left(\mathrm{C}_{8} \mathrm{H}_{18}\right)\) to produce water vapor and carbon dioxide gas. b. Assuming gasoline is \(100 . \%\) isooctane, with a density of 0.692 \(\mathrm{g} / \mathrm{mL}\) , what is the theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\) gal of gasoline (the approximate annual consumption of gasoline in the United States)?

The unbalanced chemical equation for the combustion of isooctane to form water vapor and carbon dioxide gas is: \(C_8H_{18}(l) + O_2(g) \rightarrow CO_2(g) + H_2O(g)\)

To balance the chemical equation, we will adjust the coefficients of the reactants and products accordingly: \(C_8H_{18}(l) + 12.5O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)\) Thus, the balanced chemical equation for the combustion of isooctane to produce water vapor and carbon dioxide gas is: \(C_8H_{18}(l) + 12.5O_2(g) \rightarrow 8CO_2(g) + 9H_2O(g)\) #b. Calculate the theoretical yield of carbon dioxide produced by the combustion of a given amount of gasoline#

We are given the following information: - Gasoline volume = \(1.2 \times 10^{10}\) gal - Gasoline density = 0.692 g/mL - 1 gal = 3.78541 L = 3785.41 mL First, we convert the gasoline volume from gallons to milliliters: \(1.2 \times 10^{10}\ gal \times \frac{3.78541 \times 10^3 \ mL}{1 \ gal} = 4.542 \times 10^{13}\ mL\) Now, we convert the gasoline volume in milliliters to grams using the gasoline density: \(4.542 \times 10^{13}\ mL \times \frac{0.692 \ g}{1 \ mL} = 3.1414 \times 10^{13} \ g\)

Based on the balanced chemical equation, 1 mol of \(C_8H_{18}\) produces 8 mol of \(CO_2\). We will use stoichiometry to calculate the mass of \(CO_2\) produced: Molar mass of \(C_8H_{18}\) = 8(12.01 g/mol C) + 18(1.008 g/mol H) = 114.23 g/mol Moles of \(C_8H_{18}\) = \(\frac{3.1414 \times 10^{13} \ g}{114.23 \ g/mol}\) = \(2.7482 \times 10^{11}\ mol\) Moles of \(CO_2\) produced = 8 mol \(CO_2\) × \(2.7482 \times 10^{11}\ mol \, C_8H_{18}\) = \(2.1986 \times 10^{12}\ mol\) Molar mass of \(CO_2\) = 12.01 g/mol C + 2(16.00 g/mol) = 44.01 g/mol Mass of \(CO_2\) produced = \(2.1986 \times 10^{12}\ mol \times \frac{44.01 \ g}{1 \ mol} = 9.6787 \times 10^{13} \ g\)

We are asked to find the theoretical yield of carbon dioxide in the form of "amount of gallons consumed" which is given in the problem. So, we need to convert the mass of \(CO_2\) back to the volume of gasoline: Mass of \(CO_2\) produced = 9.6787 × 10¹³ g Gasoline density = 0.692 g/mL Molar mass of \(C_8H_{18}\) = 114.23 g/mol Molar mass of \(CO_2\) = 44.01 g/mol So, 1 g of gasoline produces approximately \(8 \times \frac{1}{0.692} \times \frac{44.01}{114.23} \approx 4.353\) g of \(CO_2\). Now, we can find the volume of gasoline that produces the calculated mass of \(CO_2\): Volume of gasoline consumed = \(\frac{9.6787 \times 10^{13} \ g}{4.353 \ g/mL} = 2.2226 \times 10^{13} \ mL\) Finally, we convert the volume of gasoline to gallons: \(2.2226 \times 10^{13} \ mL \times \frac{1 \ gal}{3.78541 \times 10^3\ mL} = 5.8706 \times 10^{9} \ gal\) Thus, the theoretical yield of carbon dioxide produced by the combustion of \(1.2 \times 10^{10}\ gal\) of gasoline (the approximate annual consumption of gasoline in the United States) is approximately \(5.87 \times 10^9 \ gal\).