50 kg of $N_{2} (g)$ and 10 kg of $H_{2}$ (g) are mixed to produce $N H_{3} (g)$ . The correct statements about the reaction is ?

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dinitrogen is limiting reagent

dihydrogen is limiting reagent

56.6 (approx) kg of $N H_{3} (g)$ is formed

56.6(approx) kg of $H_{2}$ (g) is consumed

answer is C, B.

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Detailed Solution

$\begin{array}{l} N_{2 (g)} + 3 H_{2 (g)} \to 2 N H_{3} (g) \\ m o l e s = \frac{50, 000}{28} \frac{10, 000}{2} \\ 1785 m o l e s 5, 000 m o l e s \\ \frac{1785}{1} \frac{5, 000}{3} \\ = 1785 1666.6 (l e s s e r) \\ (e x c e s s r e a g e n t) (∴ L i m i t i n g r e a g e n t) \end{array}$

Amount of product formed is calculated as per limiting reagent

$\begin{array}{l} 3 m o l e s H_{2} \to 2 m o l e s N H_{3} \\ 5, 000 m o l e s \to ? \\ = \frac{5000 \times 2}{3} = 3333.33 m o l e s \\ = 3333.3 \times 17 g r \\ = 56.666 g r \\ = 56.6 k g \end{array}$