Consider the fission reaction $${ }_{0}^{1} n+{ }_{92}^{235} \mathrm{U} \longrightarrow{ }_{3}{ }_{3}^{89} \mathrm{Rb}+{ }_{55}^{144} \mathrm{Ce}+3_{-1}^{0} e+3{ }_{0}^{1} n$$ How many liters of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}\), the primary component of gasoline, must be burned to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g)\) to produce as much energy as the fission of one gram of U-235 fuel? Octane has a density of \(0.703 \mathrm{~g} / \mathrm{mL} ;\) its heat of formation is \(-249.9 \mathrm{~kJ} / \mathrm{mol}\).

To find the energy produced in the fission of 1 gram of U-235, we will use the famous equation that relates energy (E) with mass (m) and the speed of light (c): \(E=mc^2\). Since we are given the mass of U-235 (1 gram), we can easily convert it into energy using this equation. We need to convert the mass to kg before using the equation: 1 gram = 0.001 kg Speed of light, c = \(3 \times 10^8 \: \mathrm{m/s}\) Now, calculate the energy produced: \(E = (0.001 \, \text{kg}) \times (3 \times10^8 \: \mathrm{m/s})^2\) \(E = 9 \times 10^{13} \: \mathrm{J}\)

We are given the heat of formation of octane, which is \(-249.9 \, \mathrm{kJ/mol}\). This means that 249.9 kJ of energy is released when 1 mole of octane is burned.

Now, we need to find how many moles of octane must be burned to produce \(9 \times 10^{13} \: \mathrm{J}\) of energy. We can use the following proportion: \(\frac{\text{moles of octane}}{249.9 \times 10^3 \, \mathrm{J}} = \frac{1 \, \text{g of U-235}}{9 \times 10^{13} \, \mathrm{J}}\) Solving for moles of octane: \(\text{moles of octane} = \frac{249.9 \times 10^3 \, \mathrm{J}}{9 \times 10^{13} \, \mathrm{J}}\) \(\text{moles of octane} \approx 2.777 \times 10^{-12}\)

We need to convert the moles of octane required to liters, as given in the problem. Use the density of octane and its molecular weight: \(\text{Density of octane} = 0.703\, \mathrm{g/mL}\) \(\text{Molecular weight of octane} (#_\textrm{C} = 8, #_\textrm{H} = 18)\) = \(8 (12.01 \, \text{g/mol}) + 18 (1.008 \, \text{g/mol}) \approx 114.23 \, \mathrm{g/mol}\) First, find the mass of octane in grams: \(\text{Mass of octane} = (\text{moles of octane}) \times (\text{molecular weight})\) \(\text{Mass of octane} = (2.777 \times 10^{-12} \, \mathrm{mol}) \times (114.23 \, \mathrm{g/mol})\) \(\text{Mass of octane} \approx 3.173 \times 10^{-10} \, \mathrm{g}\) Now, convert the mass of octane to volume using the density: \(\text{Volume of octane} = \frac{\text{mass}}{\text{density}}\) \(\text{Volume of octane} = \frac{3.173 \times 10^{-10} \, \mathrm{g}}{0.703 \, \mathrm{g/mL}}\) \(\text{Volume of octane} \approx 4.516 \times 10^{-10} \, \mathrm{mL}\) Convert mL to L: \(\text{Volume of octane} = 4.516 \times 10^{-13} \, \mathrm{L}\) Therefore, around \(4.516 \times 10^{-13} \, \mathrm{L}\) of octane must be burned to produce the same amount of energy as the fission of 1 gram of U-235 fuel.