Ammonia is a common ingredient in fertilizers. How much nitrogen is required to produce $100 k g$ of ammonia in the following unbalanced reaction?
$N_{2 (g)} + H_{2 (g)} \to N H_{3 (g)}$

0.0823 k g

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54.9 k g

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82.3 k g

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164.7 k g

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Solution

The correct option is B $54.9 k g$
According to Haber's process:
$N_{2} + 3 H_{2} ⇌ 2 N H_{3}$
Given, mass of ammonia $= 100 \times 1000 g r a m s = 10^{5} g r a m s$
No. of moles of ammonia $= \frac{10^{5}}{17} = 5882.35 m o l e s$
$1 m o l e$ of Nitrogen produce $3 m o l e s$ of Nitrogen
$1 m o l e N_{2} \to$ $3 m o l e s N_{2}$
$x \to 5882.32.35$
$\Rightarrow x = \frac{5882.3}{3} = 1960.7 m o l e s$
Mass of Nitrogen required $= (1960.7) \times (28) = 54901 g = 54.9 k g$

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