80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a close… | JEE Main 2026 Chemistry PYQ

MCQMediumJEE Main 2026 · 21 January, Shift 1Stoichiometry & Calculations

80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to $273 \, \text{K}$ occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is:

A
$C_2H_4$
B
$C_2H_6$
C
$C_2H_2$
D
$C_4H_{10}$

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