JEE Main 2026 · 2 Apr · Shift 2 · Chemistry Q51

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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 $\ce{CO2(g)}$ 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 and worked solution

Official answer: B

Formula. C : H = 12 : 1 by mass means 1 : 1 by moles; with two carbons the hydrocarbon is $\ce{C2H2}$ (26 g mol⁻¹).

Combustion. $\frac{3.38}{26}=0.13$ mol gives $0.26$ mol $\ce{CO2}$ $=0.26\times44=11.44$ g.

(B)

Topics:Empirical and molecular formula Stoichiometry

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JEE Main 2026 · 2 Apr · Shift 2, Chemistry question 51