Ratio of mass percentage of C : H = 12:1 Ratio of moles of C : H = 1212:11=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 = 263.38=0.13 mol The combustion reaction is: C2H2+25O2→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=11.44 g Answer: 11.44
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…
JEE Main 2026 — Chemistry Physical Chemistry
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)
Held on 2 Apr 2026 · Verified 1 Jul 2026.
Options
- A
5.68
- B
11.44
- C
22.74
- D
17.05
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