Question

In Haber's process, 50.0 g of $N_2\left(g\right)$ and 10.0g of $H_2\left(g\right)$ are mixed to produce $N{H}_3\left(g\right)$. What is 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

The correct option is A 3.33

$N_2+3H_2\rightleftharpoons 2N{H}_3$

Moles of $N_2=\frac{50}{28}=1.785moles$

Moles of $H_2=\frac{10}{2}=5moles$

1 mole of $N_2$ reacts with 3 moles $H_2$.

$H_2$ is limiting reagent.

3 moles $H_2$ form 2 moles of $N{H}_3$

Therefore, 5 moles of $H_2$ will produce $\frac{2}{3}\times 5$

$=3.33moles$

Hence, option A is correct.