40 ml of gaseous mixture of CO , CH 4 and Ne was exploded with 10 ml of oxygen on cooling, the gaseous occupied 36.5 ml . After treatment with KOH the volume reduced by 9 ml and again on treatment w

Question

40 ml of gaseous mixture of CO , CH 4 ​ and Ne was exploded with 10 ml of oxygen on cooling, the gaseous occupied 36.5 ml . After treatment with KOH the volume reduced by 9 ml and again on treatment with alkaline pyrogallol the volume after with alkaline pyrogallol the volume further reduced. Percentage of CH 4 ​ in the original mixture is

TestHub · Accepted Answer

Question Explanation: Deducing mixture composition using selective absorbents. Concept: Eudiometry.

Solution: Let initial volumes be $V_{CO} = x, V_{CH_{4}} = y$ , and $V_{Ne} = 40 - x - y$ . The volume reduction by KOH gives the volume of $CO_{2}$ produced. The further reduction by pyrogallol indicates the presence of unreacted $O_{2}$ . With this data, we can form equations to solve for x and y .

Calculation:

CO + $\frac{1}{2}$ O₂ → CO₂

$x$ $\frac{x}{2}$ $x$

CH₄ + 2O₂ → CO₂ + 2H₂O

$y$ $2 y$ $y$

$V_{C O_{2}} = x + y = 9 m l .. (1)$

$V_{f} = V_{C O_{2}} + V_{N e} - (V_{O_{2}})_{l e f t} = 36.5 m l$

$(x + y) + (40 - x - y) + (10 - \frac{x}{2} - 2 y) = 36.5$

$x + 4 y = 27 \dots\dots (2)$

by (1) & (2)

x = 3, y = 6

% of CH₄ = $\frac{6}{40} \times 100 = 15%$ .

Final Answer: 15

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