80 mL of a hydrocarbon on mixing with 264 mL of oxygen ... | null (Chemistry)

JEE Mains 2026 21st January Shift 1

Chemistry

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 K occupy 224 mL . When the system is treated with KOH solution, the volume decreases to 64 mL . The formula of the hydrocarbon is :

\mathrm{C}_{2}\,\mathrm{H}_{4}

\mathrm{C}_{4}\,\mathrm{H}_{10}

\mathrm{C}_{2}\,\mathrm{H}_{2}

\mathrm{C}_{2}\,\mathrm{H}_{6}

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Explanation

\mathrm{C}_{\mathrm{x}} \mathrm{H}_{\mathrm{y}(\mathrm{~g})}+\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{xCO}_{2(\mathrm{~g})}+\frac{\mathrm{y}}{2} \mathrm{H}_{2} \mathrm{O}_{(\ell)}\\

\begin{array}{llccl}\mathrm{t}=0 & 80 & 264 & 0 & - \\ \mathrm{t}=\mathrm{t}_{\text {final }} & - & 264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right) & 80 \mathrm{x} & -\end{array}\\

264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)+80 \mathrm{x}=224\\

264-\frac{80 \mathrm{y}}{4}=224\\

40=\frac{80 \mathrm{y}}{4} \Rightarrow \mathrm{y}=2\\

264-80\left(\mathrm{x}+\frac{\mathrm{y}}{4}\right)=64\\

264-80\left(\mathrm{x}+\frac{1}{2}\right)=64\\

264-80 \mathrm{x}-40=64\\

\mathrm{x}=2

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