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What volume of oxygen gas (in litres) at 1 atm and 273 K is required to completely combust 1.97 g of propane (C3H8)?

1. 5
2. 10
3. 0.5
4. 2

Correct answer: 5

Solution

Balanced equation: C3H8 + 5O2 -> 3CO2 + 4H2O. Molar mass of C3H8 = 3*12 + 8*1 = 44 g/mol. Moles of C3H8 = 1.97/44 = 0.04477 mol ~ 1/22.4 mol. Wait: 1.97/44 = 0.04477. Moles of O2 = 5 * 0.04477 = 0.2239 mol. Volume at STP (273 K, 1 atm): V = 0.2239 * 22.4 = 5.01 L ~ 5 L.

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