Final answer:
To completely combust 360 g of C₅H₁₂, we calculate the moles of C₅H₁₂ and moles of O₂ needed. At STP, 896 L of O₂ is required for combustion, and since air is 20% O₂, we need a total of 4480 L of air. The correct option is C.
Step-by-step explanation:
The question concerns how much volume of air is required for the complete combustion of 360 g of pentane (C₅H₁₂) at 1 atm and 273 K. First, the balanced chemical equation for the combustion of pentane must be determined:
C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O
One mole of C₅H₁₂ requires 8 moles of O₂ for complete combustion. Using the molar mass of C₅H₁₂, which is 72.2 g/mol, we can calculate the moles of C₅H₁₂:
360 g C₅H₁₂ × (1 mole/72.2 g) = 5 moles C₅H₁₂
Since one mole of pentane needs 8 moles of O₂, 5 moles of pentane will need:
5 moles C₅H₁₂ × 8 moles O₂/mole C₅H₁₂ = 40 moles O₂
At STP (Standard Temperature and Pressure), one mole of any gas occupies 22.4 liters, so:
40 moles O₂ × 22.4 L/mole = 896 L O₂
Considering that air is 20% O₂ by volume, the total volume of air needed is:
896 L O₂ ÷ 0.20 = 4480 L air
The correct answer for the given problem is 4480 L of air.