How many liters of air \(\left(78\right.\) percent \(\mathrm{N}_{2}, 22\) percent \(\mathrm{O}_{2}\) by volume) at \(20^{\circ} \mathrm{C}\) and 1.00 atm are needed for the complete combustion of \(1.0 \mathrm{~L}\) of octane, \(\mathrm{C}_{8} \mathrm{H}_{18}, \mathrm{a}\) typical gasoline component that has a density of \(0.70 \mathrm{~g} / \mathrm{mL} ?\)

The general reaction for a combustion reaction is \( \mathrm{Hydrocarbon} + \mathrm{O}_{2} \rightarrow \mathrm{CO}_{2} + \mathrm{H}_{2}\mathrm{O} \). For octane \(\mathrm{C}_{8}\mathrm{H}_{18}\), this becomes \( \mathrm{C}_{8}\mathrm{H}_{18} + x \cdot \mathrm{O}_{2} \rightarrow y \cdot \mathrm{CO}_{2} + z \cdot \mathrm{H}_{2}\mathrm{O} \), Balancing this equation, we find \( \mathrm{C}_{8}\mathrm{H}_{18} + 12.5 \cdot \mathrm{O}_{2} \rightarrow 8 \cdot \mathrm{CO}_{2} + 9 \cdot \mathrm{H}_{2}\mathrm{O} \)

Octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\)) has a volume of 1.0 L and a density of 0.70 g/mL. We can convert octane's volume to its mass using the density (mass = volume x density). So, the mass of octane is \( 1.0 \, \mathrm{L} \times 1000 \, \mathrm{mL/L} \times 0.70 \, \mathrm{g/mL} = 700 \, \mathrm{g} \).We know from the balanced chemical equation that one mole of octane (\(\mathrm{C}_{8}\mathrm{H}_{18}\)) needs 12.5 moles of O2 for a complete combustion. The molecular weight of octane is approximately 114 g/mol. So, the moles of octane are \( 700 \, \mathrm{g} /\, 114 \, \mathrm{g/mol} = 6.14 \, \mathrm{mol} \).Therefore, the moles of O2 needed = \( 6.14 \, \mathrm{mol} octane \times 12.5 \, \mathrm{mol} O_{2} /\, \mathrm{mol} \, \mathrm{octane} = 76.75 \, \mathrm{mol} \, \mathrm{O}_{2} \).

From the ideal gas law, we know that 1 mole of an ideal gas at 1 atm and \(20^{\circ}\) C or \(293 \, \mathrm{K}\) occupies approximately 24.5 L. This gives the volume of O2 required for the combustion as \(76.75 \, \mathrm{mol} \times 24.5 \, \mathrm{L/mol} = 1880.125 \, \mathrm{L}\)

Since only 22% of air by volume is composed of \(\mathrm{O}_{2}\), the total volume of air required can be calculated by dividing the volume of O2 by the volume percentage of O2 in air. So, Volume of air = \( \frac{1880.125 \, \mathrm{L}}{0.22} \approx 8545.1 \, \mathrm{L} \).