(a) If an automobile travels 225 mi with a gas mileage of 20.5 mi/gal, how many kilograms of \(\mathrm{CO}_{2}\) are produced? Assume that the gasoline is composed of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}(l),\) whose density is 0.692 \(\mathrm{g} / \mathrm{mL}\) . (b) Repeat the calculation for a truck that has a gas mileage of 5 \(\mathrm{mi} / \mathrm{gal} .\)

First, let's convert the miles into gallons of fuel consumed for both automobile and truck. For the automobile: 225 mi / 20.5 mi/gal = \(10.976\) gallons of octane consumed For the truck: 225 mi / 5 mi/gal = \(45\) gallons of octane consumed

Now let's convert the gallons of octane consumed into moles. First, we need to convert gallons to liters and then to milliliters (1 gal = 3.785 L and 1 L = 1000 mL). Now the volume can be converted to mass based on the given density (0.692 g/mL): The molar mass of octane C8H18 is: \(8 * 12.01 + 18 * 1.008 = 114.23\) g/mol For the automobile: \((10.976 \text{ gal})\left(\frac{3.785\text{L}}{1\text{ gal}}\right)\left(\frac{1000\text{mL}}{1\text{ L}}\right)\left(\frac{0.692 \text{g}}{1\text{ ml}}\right)\left(\frac{1 \text{ mol of octane}}{114.23\text{g}}\right) = 274.36 \text{ mol of octane} \) For the truck: \((45 \text{ gal})\left(\frac{3.785\text{L}}{1\text{ gal}}\right)\left(\frac{1000\text{mL}}{1\text{ L}}\right)\left(\frac{0.692 \text{g}}{1\text{ ml}}\right)\left(\frac{1 \text{ mol of octane}}{114.23\text{g}}\right) = 1,123.84 \text{ mol of octane} \)

Looking at the balanced chemical equation for the combustion of octane: \(2C_8H_{18} + 25O_2 \rightarrow 16CO_2 + 18H_2O\) Moles of CO2 produced = 8 times moles of octane consumed For the automobile: \(274.36 \text{ mol of octane} * 8 = 2,195 \text{ mol of } CO_2\) For the truck: \(1,123.84 \text{ mol of octane} * 8 = 8,990.72 \text{ mol of } CO_2\)

Now, we will convert the moles of CO2 to the mass in kilograms. The molar mass of CO2 is: \(12.01 + 2 * 16.0 = 44.01\) g/mol For the automobile: \((2,195 \text{ mol } CO_2)\left(\frac{44.01\text{g}}{1\text{ mol }}\right)\left(\frac{1 \text{ kg }}{1000\text{g}}\right) = 96.62 \text{ kg of } CO_2\) For the truck: \((8,990.72 \text{ mol } CO_2)\left(\frac{44.01\text{g}}{1\text{ mol }}\right)\left(\frac{1 \text{ kg }}{1000\text{g}}\right) = 395.71 \text{ kg of } CO_2\) So, if an automobile travels 225 mi with a gas mileage of 20.5 mi/gal, it produces 96.62 kg of CO2, and if a truck travels the same distance with a gas mileage of 5 mi/gal, it produces 395.71 kg of CO2.