JEE Main 2026 — Apr 2, Shift 2 — Chemistry solution

JEE Main2026Apr 2, Shift 2Chemistry

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 : $16$ )

A
$5.68$
B
$11.44$
C
$22.74$
D
$17.05$

Solution

Mass ratio C : H = $12 : 1$ implies mole ratio C : H = $1 : 1$ . With $2$ carbon atoms, the molecular formula is C $_{2}$ H $_{2}$ (molar mass $= 26$ g/mol).

Moles of C $_{2}$ H $_{2} = \frac{3.38}{26}$ mol.

Combustion: C $_{2}$ H $_{2} + \frac{5}{2}$ O $_{2} \to 2$ CO $_{2} +$ H $_{2}$ O. Each mole of acetylene produces $2$ moles of CO $_{2}$ .

Mass of CO $_{2} = \frac{3.38}{26} \times 2 \times 44 = 11.44$ g.

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