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 CO_2(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)
B. 11.44
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 C_2H_2. Molar mass of C_2H_2 = 2 12 + 2 1 = 26 g mol^ -1 Moles of C_2H_2 in 3.38 g = 3.38 26 = 0.13 mol The combustion reaction is: C_2H_2 + 5 2 O_2 2CO_2 + H_2O From the stoichiometry, 1 mole of C_2H_2 produces 2 moles of CO_2. Moles of CO_2 produced = 2 0.13 = 0.26 mol Molar mass of CO_2 = 12 + 2 16 = 44 g mol^ -1 Weight of CO_2 formed = 0.26 44 = 11.44 g Answer: 11.44
Related: Chemistry — Some Basic Concepts of Chemistry · All PYQ Banks