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 )

Chemistry · Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

JEE Main 2026 — 2 April, Evening Shift — Question 48

The ratio of mass percentage $(\mathrm{w} / \mathrm{w})$ of $\mathrm{C}: \mathrm{H}$ in a hydrocarbon is $12: 1$ . It has two carbon atoms. The weight (in g ) of $\mathrm{CO}_{2}(\mathrm{~g})$ formed when 3.38 g of this hydrocarbon is completely burnt in oxygen is :(Given : Molar mass in $\mathrm{g} \mathrm{mol}^{-1} \mathrm{C}: 12, \mathrm{H}: 1, \mathrm{O}: 16$ )

Option A:
5.68
Option B:
11.44
Correct
Option C:
22.74
Option D:
17.05

Answer: B

Step-by-step solution

Molecular formula of hydrocarbon is $\mathrm{C}_{2} \mathrm{H}_{2}$ Moles of $\mathrm{C}_{2} \mathrm{H}_{2}=\frac{3.38}{26}$ $\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{~g})+\frac{5}{2} \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(l)$ Moles of $\mathrm{CO}_{2}$ produced $=\left(\frac{3.38}{26} \times 2\right)$ Mass of $\mathrm{CO}_{2}$ produced $=\left(\frac{3.38}{26} \times 2\right) \times 44=11.44$

Answer key and solution verified before publishing.

Practise Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 2 April, Evening Shift
Subject: Chemistry
Chapter: Some Basic Concepts of Chemistry (Mole Concept) + Stoichiometry
Topic: Percentage composition & Empirical Formula