# Dinitrogen and dihydrogen react with each other

Question.

Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation

$\mathrm{N}_{2}(g)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)$

(i) Calculate the mass of ammonia produced if $2.00 \times 10^{3} \mathrm{~g}$ dinitrogen reacts with $1.00$ $\times 10^{3} \mathrm{~g}$ of dihydrogen.

(ii) Will any of the two reactants remain unreacted?

(iii) If yes, which one and what would be its mass?

Solution:

(i)Balancing the given chemical equation,

$\mathrm{N}_{2(g)}+3 \mathrm{H}_{2(g)} \rightarrow 2 \mathrm{NH}_{3(g)}$

From the equation, 1 mole $(28 \mathrm{~g})$ of dinitrogen reacts with 3 mole $(6 \mathrm{~g})$ of dihydrogen to give 2 mole $(34 \mathrm{~g})$ of ammonia.

$\Rightarrow 2.00 \times 10^{3} \mathrm{~g}$ of dinitrogen will react with $\frac{6 \mathrm{~g}}{28 \mathrm{~g}} \times 2.00 \times 10^{3} \mathrm{~g} \quad$ dihydrogen

i.e., $2.00 \times 10^{3} \mathrm{~g}$ of dinitrogen will react with $428.6 \mathrm{~g}$ of dihydrogen.

Given,

Amount of dihydrogen $=1.00 \times 10^{3} \mathrm{~g}$ Hence,

$\mathrm{N}_{2}$ is the limiting reagent.

$\therefore 28 \mathrm{~g}$ of $\mathrm{N}_{2}$ produces $34 \mathrm{~g}$ of $\mathrm{NH}_{3}$

Hence, mass of ammonia produced by $2000 \mathrm{~g}$ of $\mathrm{N}_{2}=\frac{34 \mathrm{~g}}{28 \mathrm{~g}} \times 2000 \mathrm{~g}$

$=2428.57 \mathrm{~g}$

(ii) $\mathrm{N}_{2}$ is the limiting reagent and $\mathrm{H}_{2}$ is the excess reagent. Hence, $\mathrm{H}_{2}$ will remain unreacted.

(iii) Mass of dihydrogen left unreacted $=1.00 \times 10^{3} \mathrm{~g}-428.6 \mathrm{~g}$

$=571.4 \mathrm{~g}$