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 $\mathrm{g~mol}^{-1}$ C: 12, H: 1, O: 16)

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Some Basic Concepts of 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

CO X_{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)

Answer: (B)

A carbon-to-hydrogen mass ratio of

12 : 1

with two carbons fixes the masses as

24

for carbon and

2

for hydrogen, that is two hydrogen atoms, so the hydrocarbon is

C X_{2} H X_{2}

of molar mass

26 g mo l^{- 1}

. Complete combustion follows

C X_{2} H X_{2} + \frac{5}{2} O X_{2} 2 CO X_{2} + H X_{2} O

, yielding two moles of

CO X_{2}

per mole of acetylene. Hence

m_{CO X_{2}} = \frac{3.38}{26} \times 2 \times 44 = 11.44 g .

The mass of carbon dioxide formed is

11.44

JEE Main · 2026 · 02 Apr 2026 · Shift 2 · Single correct option