A mixture containing 28 g of ethene ( C ₂ H ₄ ) and 48 g of oxygen gas is allowed to react in a closed vessel. The mass of carbon dioxide produced after the reaction is completed is : (Given molar masses in g mol ⁻¹ : C = 12 , H = 1 , O = 16 )

Step-by-step solution

The balanced chemical equation for the combustion of ethene is: C ₂ H ₄( g ) + 3 O ₂( g ) 2 CO ₂( g ) + 2 H ₂ O ( l ) Molar mass of C ₂ H ₄ = 28 g mol ⁻¹ Moles of C ₂ H ₄ = 28 28 = 1 mol Molar mass of O ₂ = 32 g mol ⁻¹ Moles of O ₂ = 48 32 = 1.5 mol From the stoichiometry, 1 mol of C ₂ H ₄ requires 3 mol of O ₂ . Since only 1.5 mol of O ₂ is available, O ₂ is the limiting reagent. According to the balanced equation, 3 mol of O ₂ produces 2 mol of CO ₂ . Therefore, 1.5 mol of O ₂ will produce 2 3 1.5 = 1.0 mol of CO