The mole fraction of oxygen molecules in dry air is 0.2095. What volume of dry air at 1.00 atm and 25°C is required for burning completely 1.00 L of octane (C₈H₁₈, density = 0.7025 g/mL), yielding carbon dioxide and water? (R = 0.08206 L • atm/K • mol)

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Quizplus · Accepted Answer

First, calculate the moles of octane: 1.00 L × 0.7025 g/mL = 702.5 g. Molar mass of C8H18 = 114.22 g/mol, so moles of octane = 702.5 g / 114.22 g/mol = 6.15 mol. The combustion reaction is 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O, so 6.15 mol C8H18 requires 6.15 × (25/2) = 76.875 mol O2. Using PV = nRT, V = nRT/P = 76.875 mol × 0.08206 L•atm/K•mol × 298 K / 1 atm = 1878.5 L of O2. Since the mole fraction of O2 in air is 0.2095, the volume of air needed is 1878.5 L / 0.2095 = 8968.5 L, approximately 8970 L.

The Mole Fraction of Oxygen Molecules in Dry Air Is