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Multiple choice chemistry nitrogen and sulfur ammonia and fertilizers preparation of ammonia-laboratory method and haber's process ammonia

In Haber's process $50.0\ g$ of $N _{2}(g)$ and $10.0\ g$ of $H _{2}(g)$ are mixed to produced $NH _{3}(g)$. What are the number of moles of $NH _{3}(g)$ formed?

  1. $3.33$
  2. $2.36$
  3. $2.01$
  4. $5.36$
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A Correct answer
Explanation
$N _2 + 3H _2 \rightarrow2NH3$
Here.
28 g $N _2$ reacts with 6 g of $H _2$
So,
50 g of $N _2$ react with $\dfrac{6}{28}\times50 g -H _2 = 10.7 g$ but we have only 10g so $H _2$ is limiting reagent..

6 g $H _2$ gives 34g $NH _3$
10g $H _2$ will give $\dfrac{34}{6}\times10=56.67g -NH _3$
So number of moles of $NH _3$ formed is,
$=\dfrac{Given-mass}{Molar-mass}=\dfrac{56.67}{17}=3.33-moles$