Question

In Haber's process $50.0g$ of $N_2\left(g\right)$ and $10.0g$ of $H_2\left(g\right)$ are mixed to produced $N{H}_3\left(g\right)$. What are the number of moles of $N{H}_3\left(g\right)$ formed?

• A. $3.33$
• B. $2.36$
• C. $2.01$
• D. $5.36$

Answer 1

Answer: (A)

$3.33$

$N_2+3H_2\to 2NH3$

Here.

28 g $N_2$ reacts with 6 g of $H_2$

So,

50 g of $N_2$ react with $\frac{6}{28}\times 50g-H_2=10.7g$ but we have only 10g so $H_2$ is limiting reagent..

6 g $H_2$ gives 34g $N{H}_3$

10g $H_2$ will give $\frac{34}{6}\times 10=56.67g-N{H}_3$

So number of moles of $N{H}_3$ formed is,

$=\frac{Given-mass}{Molar-mass}=\frac{56.67}{17}=3.33-moles$