Question

$50kg{N}_2$ and $10kg{H}_2$ are mixed to produce ${NH}_3$. Calculate the ammonia gas formed. Identify the limiting reactant.

Answer 1

$N_2+3H_2⟶2N{H}_3$

Molecular mass of Nitrogen $=28g/mol=0.028kg/mol$

Molecular mass of Hydrogen $=6g/mol=0.006kg/mol$

Molecular mass of Ammonia $=17g/mol=0.017kg/mol$

Now, according to the balanced chemical equation,

0.028 kg of Nitrogen reacts with 0.006 kg of Hydrogen.

$\therefore$ 50 kg of Nitrogen reacts with $\left[\frac{(0.006\times 50)}{0.028}\right]=10.71kg$ of Hydrogen.

The amount of Hydrogen (given 10 kg) is less than the amount required (i.e., 10.71 kg) for 50 kg of Nitrogen.

Therefore, Hydrogen is the limiting reagent.

Hence, the formation of Ammonia will depend on the amount of Hydrogen available for reaction.

$\because$ Amount of ammonia produced by 0.006 kg of hydrogen $=2\times 0.017=0.034kg$

$\therefore$ Amount of ammonia produced by 10 kg of Hydrogen $=\frac{0.034\times 10}{0.006}=56.67kg$

Hence, 56.67 kg of ammonia gas will be formed and hydrogen will be limiting reagent.