The ratio of mass percentage (w/w) of C : H in a hydrocarbon is 12 : 1 . It has two carbon atoms. The weight (in g) of CO₂(g) formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is : (Given : Molar mass in g mol ⁻¹ C : 12, H : 1, O : 16)

Step-by-step solution

Ratio of mass percentage of C : H = 12 : 1 Ratio of moles of C : H = 12 12 : 1 1 = 1 : 1 The empirical formula is CH . Since the hydrocarbon has two carbon atoms, its molecular formula is C₂H₂ . Molar mass of C₂H₂ = 2 12 + 2 1 = 26 g mol ⁻¹ Moles of C₂H₂ in 3.38 g = 3.38 26 = 0.13 mol The combustion reaction is: C₂H₂ + 5 2 O₂ 2CO₂ + H₂O From the stoichiometry, 1 mole of C₂H₂ produces 2 moles of CO₂ . Moles of CO₂ produced = 2 0.13 = 0.26 mol Molar mass of CO₂ = 12 + 2 16 = 44 g mol ⁻¹ Weight of CO₂ formed = 0.26 44