Some Basic Concepts of Chemistry - Result Question 36
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37. A $20.0 \mathrm{~cm}^3$ mixture of $\mathrm{CO}, \mathrm{CH}_4$ and He gases is exploded by an electric discharge at room temperature with excess of oxygen. The volume contraction is found to be $13.0 \mathrm{~cm}^3$.
A further contraction of $14.0 \mathrm{~cm}^3$ occurs when the residual gas is treated with KOH solution. Find out the composition of the gaseous mixture in terms of volume percentage.
(1995,4 M)
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Solution:
- The reaction involved in the explosion process is
$\begin{aligned} & \mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow \mathrm{CO}_2(\mathrm{~g}) \\ & x \mathrm{~mL} \quad \frac{x}{2} \mathrm{~mL} \hspace{15mm} x \mathrm{~mL} \\ \end{aligned}$
$\underset{y \mathrm{~mL}}{\mathrm{CH}_4(g)}+\underset{2 y \mathrm{~mL}}{2 \mathrm{O}_2(g)} \longrightarrow \underset{y \mathrm{~mL}}{\mathrm{CO}_2(g)}+2 \mathrm{H}_2 \mathrm{O}(l)$
The first step volume contraction can be calculated as:
$\left(x+\frac{x}{2}+y+2 y\right)-(x+y)=13$
$\Rightarrow \quad x+4 y=26 \hspace{15mm} …(i)$
The second volume contraction is due to absorption of $\mathrm{CO}_2$.
Hence, $\hspace{20mm} x+y=14 \hspace{15mm} …(ii)$
Now, solving equations (i) and (ii), $x=10 \mathrm{~mL}, y=4 \mathrm{~mL}$ and volume of $\mathrm{He}=20-14=6 \mathrm{~mL}$
$\begin{aligned} & \Rightarrow \quad \mathrm{Vol} % \text { of } \mathrm{CO}=\frac{10}{20} \times 100=50 % \\ & \mathrm{Vol} % \text { of } \mathrm{CH}_4=\frac{4}{20} \times 100=20 % \\ & \text { Vol } % \text { of } \mathrm{He}=30 % \\ \end{aligned}$
BETA
