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Consider the reaction $N_{2} (g) + 3 H_{2} (g) \to 2 {NH}_{3} (g)$
$n_{0} 1.5 mol 1.5 mol$
with the progress of reaction it is found that there is a formation of 0.8 mol of ${NH}_{3}$ At this stage, the extent of reaction is

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$0.2 mol$

$0.3 mol$

$0.4 mol$

$0.5 mol$

answer is C.

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Detailed Solution

We have $ξ = \frac{Δ n_{{NH}_{3}}}{v_{{NH}_{3}}} = \frac{0.8 mol}{2} = 0.4 mol$

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