Question

5. A reaction mixture for the production of
NH, gas contains 250g of N, gas and 50 g
of H, gas under suitable conditions. Iden-
tify the limiting reactant, if any and cal-
culate the mass of NH, gas produced.​

Answer 1

Answer: $H_2$ is the limiting reagent and 283.9 g of $NH_3$  will be produced.

Explanation:

$N_2+3H_2\rightarrow 2NH_3$

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

$\text{Number of moles of nitrogen}=\frac{250g}{28g/mol}=8.92moles$

$\text{Number of moles of hydrogen}=\frac{50g}{2g/mol}=25moles$

According to stoichiometry:

3 moles of $H_2$ react with 1 mole of $N_2$

25 moles of $H_2$ will react with=$\frac{1}{3}\times 25=8.33moles$ of $N_2$

Thus $H_2$ is the limiting reagent as it limits the formation of product and $N_2$  is the excess reagent as (8.92-8.33)=0.59 moles are left unreacted.

3 moles of $H_2$ produces 2 mole of $NH_3$

25 moles of $H_2$ produces=$\frac{2}{3}\times 25=16.7moles$ of $NH_3$

Mass of $NH_3$ =$moles\times {\text {Molar mass}}=16.7\times 17=283.9g$

Thus $H_2$ is the limiting reagent and 283.9 g of $NH_3$  will be produced.