Question

How many grams of hydrogen gas is required to produce 4.8 mol of ammonia on a reaction with an excess of nitrogen?

• A. $17\text{g}$
• B. $7.2\text{g}$
• C. $15\text{g}$
• D. $14.4\text{g}$

Answer 1

The correct option is D $14.4\text{g}$

The balanced chemical equation is:
$N_2+3H_2\to 2N{H}_3$

From stoichiometry,
2 mol of $N{H}_3$ is produced from 3 mol of $H_2$.
4.8 mol of $N{H}_3$ will be produced from = $\frac{3}{2}\times 4.8$ = 7.2 mol of $H_2$
$\therefore$ mass of $H_2=7.2\times 2=14.4\text{g}$