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 :

### Related Formula mathrmC_mathrmxmathrmH_mathrmy(g) + left(mathrmx + fracmathrmy4right)mathrmO_2(g) longrightarrow mathrmxCO_2(g) + fracmathrmy2mathrmH_2mathrmO_(ell)$$\mathrm{C}_\mathrm{x}\mathrm{H}_\mathrm{y(g)} + \left(\mathrm{x} + \frac{\mathrm{y}}{4}\right)\mathrm{O}_{2(g)} \longrightarrow \mathrm{xCO}_{2(g)} + \frac{\mathrm{y}}{2}\mathrm{H}_2\mathrm{O}_{(\ell)}$$ ### Core Logic Let the volume of hydrocarbon be V = 80$V = 80$ mL. Initial volume of O_2 = 264$O_2 = 264$ mL. At 273 K, H_2O$H_2O$ is liquid, so its volume is neglected. Volume of CO_2$CO_2$ formed = 80x$80x$ mL. Volume of O_2$O_2$ used = 80left(x + fracy4right)$80\left(x + \frac{y}{4}\right)$ mL. Unreacted O_2$O_2$ = 264 - 80left(x + fracy4right)$264 - 80\left(x + \frac{y}{4}\right)$ mL. Total residual volume = V_CO_2 + V_unreacted \ O_2 = 224$V_{CO_2} + V_{unreacted \ O_2} = 224$ mL. 80x + 264 - 80left(x + fracy4right) = 224$$80x + 264 - 80\left(x + \frac{y}{4}\right) = 224$$ 264 - frac80y4 = 224$$264 - \frac{80y}{4} = 224$$ 40 = 20y implies y = 2$$40 = 20y \implies y = 2$$ After treatment with KOH, CO_2$CO_2$ is absorbed. The remaining volume is unreacted O_2$O_2$, which is 64 mL. 264 - 80left(x + fracy4right) = 64$$264 - 80\left(x + \frac{y}{4}\right) = 64$$ Substitute y = 2$y = 2$: 264 - 80left(x + frac12right) = 64$$264 - 80\left(x + \frac{1}{2}\right) = 64$$ 264 - 80x - 40 = 64$$264 - 80x - 40 = 64$$ 224 - 80x = 64$224 - 80x = 64$ 80x = 160 implies x = 2$$80x = 160 \implies x = 2$$ The hydrocarbon is mathrmC_2mathrmH_2$\mathrm{C}_2\mathrm{H}_2$. ### Pattern Recognition Volume decrease by KOH indicates the volume of CO_2$CO_2$ produced. V_CO_2 = 224 - 64 = 160$V_{CO_2} = 224 - 64 = 160$ mL. V_HC = 80$V_{HC} = 80$ mL. So x = frac16080 = 2$x = \frac{160}{80} = 2$. Total volume reduction = 264 - 64 = 200 mL (O_2$O_2$ consumed). O_2$O_2$ consumed = 80(x + y/4) = 200 implies 2 + y/4 = 2.5 implies y/4 = 0.5 implies y = 2$80(x + y/4) = 200 \implies 2 + y/4 = 2.5 \implies y/4 = 0.5 \implies y = 2$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Some Basic Concepts of Chemistry