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For a reaction,
$N_{2 (g)} + 3 H_{2 (g)} ⟶ 2 {N H}_{3 (g)}$
identify dihydrogen $(H_{2})$ as a limiting reagent in the following reaction mixtures

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$28 g of N_{2} + 6 g of H_{2}$

$14 g of N_{2} + 4 g of H_{2}$

$35 g of N_{2} + 8 g of H_{2}$

$56 g of N_{2} + 10 g of H_{2}$

answer is C.

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Detailed Solution

$N_{2 (g)} + 3 H_{2 (g)} ⟶ 2 {N H}_{3 (g)}$

In a given chemical reaction 1 mole of $N_{2} r e q u i r e s 3 m o l e s o f H_{2}$ for the formation of ammonia.

Mole ratio of $N_{2} a n d H_{2} i s 1 : 3$ .

$mole = \frac{weight}{molecular weight}$

So, In this proportion $56 g of N_{2} (2 m o l e s) + 10 g of H_{2} (5 m o l e s), H_{2}$ act as limiting reagent.

Therefore, option C is correct.

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