Question

The mass of ${NH}_3$ produced on reacting $28$ of $N_2$ with $4$ of $H_2$ is:

• A. $34.0$
• B. $22.67$
• C. $34.23$
• D. $17.0$

Answer 1

Answer: (B)

$22.67$

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

28 g of $N_2\equiv$ 1 mole of $N_2$

4g of $H_2$ $\equiv$ 2 moles of $H_2$

Therefore, $H_2$ is a limiting reagent

Therefore, moles of $N{H}_3$ produced $=2\times \frac{2}{3}$ moles

=$\frac{4}{3}$ moles

Mass of $N{H}_3$ produced=$\frac{4}{3}\times 17$

$=22.67$