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)
- 5.68
- 11.44
- 22.74
- 17.05
Answer
(2) 11.44
Solution
The mole ratio C : H = 1212 : 11 = 1 : 1, so the empirical formula is CH; the molecule has two C atoms, the molecular formula is C2H2 (M = 26 g mol-1).
From 2C2H2 + 5O2 4CO2 + 2H2O, 1 mol C2H2 gives 2 mol CO2, so
n(C2H2) = 3.3826 = 0.13 mol
n(CO2) = 2 0.13 = 0.26 mol
∴ Mass of CO2 = 0.26 44 = 11.44 g
Answer: (2)