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 2

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Solution

Q.
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: [2026]

1 $C_{4} H_{10}$

2 $C_{2} H_{6}$

3 $C_{2} H_{2}$

4 $C_{2} H_{4}$

Ans.
(3)

$C_{x} H_{y} {}_{(g)}+ (x + \frac{y}{4}) O_{2} {}_{(g)}⟶ x C O_{2} {}_{(g)}+ \frac{y}{2} H_{2} O_{(ℓ)}$

$\begin{matrix} t = 0 & 80 & 264 & 0 & - \\ t = t_{final} & - & 264 - 80 (x + \frac{y}{4}) & 80 x & - \end{matrix}$

$264 - 80 (x + \frac{y}{4}) + 80 x = 224$

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

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

$264 - 80 (x + \frac{y}{4}) = 64$

$264 - 80 (x + \frac{1}{2}) = 64$

$264 - 80 x - 40 = 64$

$x = 2$

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