A mixture of 2 moles of $C H_{4}$ and 34 g of $H_{2} S$ was placed in an evacuated container, which was then heated to & maintained at 727⁰C. When equilibrium is established as the gaseous reaction $C H_{4} + 2 H_{2} S ⥂ C S_{2} + 4 H_{2}$ , the total pressure in the container was 0.92 atm and the partial pressure of hydrogen was 0.2 atm.
What was the volume of container ?

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10 L

300 L

200 L

25 L

answer is B.

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Detailed Solution

$C H_{4} + 2 H_{2} S \to C S_{2} + 4 H_{2}$ Total moles = 3+2x

Initial 2 1 0 0 given $P_{H_{2}} = 0.2 = \frac{4 x}{3 + 2 x} \times 0.92$

moler At equilibrium 2-x 1-2x x 4x ∴ x = 0.183

Total moles in flask = 3+2x = 3.366

$V = \frac{n R T}{P} = 299.84 L$

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