The balanced chemical equation is $\mathrm{N}_2 + 3\mathrm{H}_2$ → $2\mathrm{NH}_3$. Moles of $\mathrm{N}_2$ = $\frac{2.00 \times 10^3}{28}$ = 71.43, Moles of $\mathrm{H}_2$ = $\frac{1.00 \times 10^3}{2}$ = 500 1 mole of $\mathrm{N}_2$ required 3 moles of $\mathrm{H}_2$ from above equation. ∴ 71.43 mole of $\mathrm{N}_2$ will require 3 × 71.43 = 214.29 mole of $\mathrm{H}_2$ But moles of $\mathrm{H}_2$ actually present = 500 moles Thus, $\mathrm{H}_2$ is in excess and will remain unreacted and $\mathrm{N}_2$ is limiting reagent. (i) 1 mole of $\mathrm{N}_2$ reacts with $\mathrm{H}_2$ to form $\mathrm{NH}_3$ = 2 moles 71.43 moles of $\mathrm{N}_2$ react with $\mathrm{H}_2$ to form $\mathrm{NH}_3$ = $\frac{2}{1}$ × 71.43 = 142.86 moles Mass of $\mathrm{NH}_3$ produced = 142.86 × 17 = 2428.6 g (ii) Hydrogen will remain unreacted. (iii) Moles of $\mathrm{H}_2$ remaining unreacted = 500 – 214.29 = 285.71 moles Mass of $\mathrm{H}_2$ left unreacted = 285.71 × 2 = 571.42 g