Question

Consider the balanced chemical equation for the combustion of methane (CH4).

mc026-1.jpg

Given that the molar mass of CO2 is 44.01 g/mol, how many liters of oxygen is required at STP to produce 88.0 g of CO2 from this reaction?
44.8 L
45.00 L
89.55 L
89.6 L

Answer 1

Balanced equation: CH4 + 2 O2 → CO2 + 2 H2O

Moles CO2 = 88.0 g / 44.01 g·mol⁻¹ = 2.00 mol CO2
From stoichiometry, O2 needed = 2 × (moles CO2) = 4.00 mol O2
At STP, 1 mol gas = 22.4 L, so V = 4.00 × 22.4 L = 89.6 L

Answer: 89.6 L.