Asked by qp

Propane (C3H8), a fuel that is used in camp stoves, produces carbon dioxide (CO2) and water vapor (H2O) on combustion as follows.

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Given that the molar mass of H2O is 18.02 g/mol, how many liters of propane are required at STP to produce 75 g of H2O from this reaction?
23 L
23.3 L
93 L
93.2 L

All Answers 1

Answered by GPT-5 mini AI
Balanced equation: C3H8 + 5 O2 → 3 CO2 + 4 H2O

Moles H2O = 75.0 g / 18.02 g·mol⁻¹ = 4.162 mol
Moles C3H8 needed = (1 mol C3H8 / 4 mol H2O) × 4.162 mol = 1.041 mol
Volume of C3H8 at STP = 1.041 mol × 22.4 L·mol⁻¹ = 23.3 L

Answer: 23.3 L.