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A $82.1 L$ container connected with mano meter contains mixture of $C S_{2} a n d H_{2} S$ gases and just the required amount of $O_{2}$ is added to form to $C O_{2}, S O_{2}$ and $H_{2} O (g)$ at $227 ° C .$ Final condition of manometer is shown. If moles of $S O_{2}$ gas produced is 0.7. Select the correct statement

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Moles of $C S_{2}$ taken 0.2

Moles of $H_{2} S$ taken 0.3

Total moles of $C O_{2} + S O_{2} + H_{2} O (g) = 1.2$

Moles of $H_{2} S$ taken 0.2

answer is A, B, C.

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Detailed Solution

$P_{g a s} + P_{H g} = P_{a t m} P_{g a s} + 304 m m = 760 m m P_{g a s} = 760 - 304 = 456 P_{g a s} = \frac{456}{760} a t m = 0.6 a t m$

Total moles of mixture (n) $= \frac{P V}{R T} = \frac{0.6 \times 82.1}{0.0821 \times 500} = 1.2$
Moles of $S O_{2} = 0.7$
Total moles of $C O_{2} + H_{2} O = 1.2 - 0.7 = 0.5$
Question Image

$\begin{array}{l} 2 x + y = 0.7 \\ x + y = 0.5 \end{array}$

Moles of $C S_{2} = 0.2$
Moles of $H_{2} S = 0.3$

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