Question
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 :
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Solution
$$\begin{matrix} & \ C_{x}H_{y(\text{ }g)} + \left( x + \frac{y}{4} \right)O_{2(\text{ }g)} \longrightarrow {xCO}_{2(\text{ }g)} + \frac{y}{2}H_{2}O_{(\mathcal{l})} \\ & t = 0\ 80\ 264\ 0 \\ & t = t_{\text{final~}}\ - \ 264 - 80\left( x + \frac{y}{4} \right)\ 80x \\ & 264 - 80\left( x + \frac{y}{4} \right) + 80x = 224 \\ & 264 - \frac{80y}{4} = 224 \\ & 40 = \frac{80y}{4} \Rightarrow y = 2 \\ & 264 - 80\left( x + \frac{y}{4} \right) = 64 \\ & 264 - 80\left( x + \frac{1}{2} \right) = 64 \\ & 264 - 80x - 40 = 64 \\ & x = 2 \end{matrix}$$
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