Some Basic Concepts of Chemistry 2 Question 37
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 $\mathrm{KOH}$ 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
$$ \begin{aligned} & \underset{x \mathrm{~mL}}{\mathrm{CO}(g)}+\frac{1}{2} \underset{\frac{x}{2} \mathrm{~mL}}{\mathrm{O}{2}(g)} \longrightarrow \underset{x \mathrm{~mL}}{\mathrm{CO}{2}(g)} \ & \mathrm{CH}{4}(g)+2 \mathrm{O}{2}(g) \longrightarrow \mathrm{CO}{2}(g)+2 \mathrm{H}{2} \mathrm{O}(l) \ & y \mathrm{~mL} \quad 2 y \mathrm{~mL} \quad y \mathrm{~mL} \end{aligned} $$
The first step volume contraction can be calculated as :
$$ \Rightarrow \quad \begin{aligned} \left(x+\frac{x}{2}+y+2 y\right)-(x+y) & =13 \ \Rightarrow \quad x+4 y & =26 \end{aligned} $$
The second volume contraction is due to absorption of $\mathrm{CO}_{2}$.
Hence, $\quad x+y=14$
Now, solving equations (i) and (ii),
$x=10 \mathrm{~mL}, y=4 \mathrm{~mL}$ and volume of $\mathrm{He}=20-14=6 \mathrm{~mL}$
$$ \Rightarrow \quad \begin{aligned} \text { Vol } % \text { of } \mathrm{CO} & =\frac{10}{20} \times 100=50 % \ \text { Vol } % \text { of } \mathrm{CH}_{4} & =\frac{4}{20} \times 100=20 % \ \text { Vol } % \text { of } \mathrm{He} & =30 % \end{aligned} $$
