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:

Question

Accepted Answer

A. 1. **Set up the general combustion equation.** Since a hydrocarbon $\mathrm{C}_x\mathrm{H}_y$ combusts as
$$
\mathrm{C}_x\mathrm{H}_y + \left(x+\frac{y}{4}\right)\mathrm{O}_2 \rightarrow x\mathrm{CO}_2 + \frac{y}{2}\mathrm{H}_2\mathrm{O}
$$
and at 273 K water is liquid (negligible volume), the residual gas after combustion consists of unreacted $\mathrm{O}_2$ plus $\mathrm{CO}_2$ produced.

2. **Use the total residual gas volume to find y.** Given 80 mL of hydrocarbon reacts with 264 mL $\mathrm{O}_2$, and the residual gas volume (unreacted $\mathrm{O}_2$ + $\mathrm{CO}_2$) is 224 mL,
$$
264 - 80\left(x+\frac{y}{4}\right) + 80x = 224
$$
Simplifying,
$$
264 - \frac{80y}{4} = 224 \quad \Rightarrow \quad 40 = 20y \quad \Rightarrow \quad y = 2
$$

3. **Use the KOH absorption step to find x.** Since KOH absorbs $\mathrm{CO}_2$, leaving only unreacted $\mathrm{O}_2$ (64 mL) behind,
$$
264 - 80\left(x+\frac{y}{4}\right) = 64
$$
Substituting $y=2$,
$$
264 - 80\left(x+\frac{1}{2}\right) = 64 \quad \Rightarrow \quad 264 - 80x - 40 = 64 \quad \Rightarrow \quad x = 2
$$

4. **Combine the results.** Since $x=2$ and $y=2$, the hydrocarbon formula is $\mathrm{C}_2\mathrm{H}_2$.

**Hence, the answer is option A — $\mathrm{C}_2\mathrm{H}_2$.**