JEE Main 2026 — 21 January, Morning Shift — Question 52

After combustion and cooling, the remaining gases are $\mathrm{CO_2}$ and unreacted $\mathrm{O_2}$. $\mathrm{KOH}$ absorbs $\mathrm{CO_2}$, thus,

Volume of $\mathrm{CO_2}$ formed $= 224 - 64 = 160\,\mathrm{mL}$

Unreacted $\mathrm{O_2} = 64\,\mathrm{mL}$

Initial $\mathrm{O_2} = 264\,\mathrm{mL}$

$\mathrm{O_2}$ consumed $= 264 - 64 = 200\,\mathrm{mL}$

Let the hydrocarbon be $\mathrm{C_x H_y}$.

$$\mathrm{C_x H_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O}$$

Gas volumes are proportional to stoichiometric coefficients.

From $\mathrm{CO_2}$ data, $\frac{160}{80} = x = 2$

From oxygen consumption, we have $\frac{200}{80} = x + \frac{y}{4} = 2.5$

$2 + \frac{y}{4} = 2.5$

$\frac{y}{4} = 0.5$

$y = 2$

Hence, the hydrocarbon is $\mathrm{C_2H_2}$.