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1 mol of N₂ is mixed with 3 mol of H₂ in a litre container. If 50% of N₂ is converted into ammonia by the reaction $N_{2} (g) + 3 H_{2} (g) ⇌ 2 {NH}_{3} (g)$ , then the total number of moles of gas at the equilibrium are

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1.5

4.5

3.0

6.0

answer is C.

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Detailed Solution

$α = 50 % = \frac{50}{100} = 0 .5 N_{2} + 3 H_{2} ⇌ 2 {NH}_{3} \begin{array}{lccc} Initial & 1 & 3 & 0 \\ Final & 1 - α & 3 - 3 α & 2 α \\ Total moles & = 1 - α + 3 - 3 α + 2 α \\ = 4 - 2 α \\ = 4 - 2 \times 0.5 = 3 \end{array}$

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