Methane gas undergoes combustion according to the following equation: CH ₄( g ) + 2 O ₂( g ) CO ₂( g ) + 2 H ₂ O ( l ) If 3.36 L of CH ₄ gas measured at STP is mixed with 1.2044 10²³ molecules of O ₂ gas, the volume of CO ₂ gas produced at STP is: (Assume molar volume of a gas at STP is 22.4 L and N_ A = 6.022 10²³ mol ⁻¹ )

Step-by-step solution

The balanced chemical equation for the combustion of methane is: CH ₄( g ) + 2 O ₂( g ) CO ₂( g ) + 2 H ₂ O ( l ) Number of moles of CH ₄ = 3.36 22.4 = 0.15 mol Number of moles of O ₂ = 1.2044 10²³ 6.022 10²³ = 0.2 mol To identify the limiting reagent, divide the moles by their stoichiometric coefficients: For CH ₄: 0.15 1 = 0.15 For O ₂: 0.2 2 = 0.1 Since 0.1 From the stoichiometry, 2 moles of O ₂ produce 1 mole of CO ₂ . Moles of CO ₂ produced = 1 2 0.2 = 0.1 mol Volume of CO ₂ produced at STP = 0.1 22.4 = 2.24 L