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Q.

Two moles of $N H_{3}$ and one mole of $H C l$ gas are taken in a container of capacity 8.21 L at 300 K to produce $N H_{4} C l (S)$ . Find total pressure (in atm) after reaction.

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Detailed Solution

$N H_{3} + H C l \to N H_{4} C l$

(g) (g) (S)

2 1 0

1 0 1

$n_{g} = n_{N H_{3}} = 1 m o l e (a f t e r r e a c t i o n)$

$P = \frac{n_{g} R T}{V} = \frac{1 \times 0.821 \times 300}{8.21} = 3 a t m$

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