$\underset{(20 g)}{N_{2} (g)} + \underset{(5 g)}{3 H_{2} (g)} ⇌ 2 N H_{3} (g)$ Consider the above reaction the limiting reagent of the reaction and number of moles of $N H_{3}$ formed respectively are

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$H_{2}, 0.71 m o l$

$N_{2}, 1.42 m o l$

$N_{2}, 0.71 m o l$

$H_{2}, 1.42 m o l$

answer is C.

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Detailed Solution

$\begin{array}{l} N_{2 (g)} + 3 H_{2 (g)} ⇌ 2 N H_{3 (g)} \\ (20 g) (5 g) \end{array}$
Moles of $N_{2} = \frac{20}{28}$ mol
Moles of $H_{2} = \frac{5}{2}$ mol
$L i m i t i n g r e a g e n t = \frac{N u m b e r o f m o l e s}{S t o i c h i o m e t r i c c o e f f i c i e n t}$
For $N H_{3} = \frac{20}{28}$
For $H_{2} = \frac{5}{2 \times 3} = \frac{5}{6}$
Clearly , $\frac{20}{28} < \frac{5}{6}$ so, $N_{2}$ is a limiting reagent.
1 mole of $N_{2}$ forms 2 mole of $N H_{3}$ .
So, $\frac{20}{28}$ mole of $N_{2}$ forms
$= 2 \times \frac{20}{28} = \frac{20}{14} = \frac{10}{7} = 1.42$ mol.