Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
$(i)$ Calculate the mass of ammonia produced if $2.00 \times 10^{3} \, g$ of dinitrogen reacts with $1.00 \times 10^{3} \, g$ of dihydrogen.
$(ii)$ Will any of the two reactants remain unreacted?
$(iii)$ If yes,which one and what would be its mass?

(N/A) $(i)$ The balanced chemical equation is:
$N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
From the equation,$1 \, \text{mol}$ $(28 \, g)$ of $N_{2}$ reacts with $3 \, \text{mol}$ $(6 \, g)$ of $H_{2}$ to produce $2 \, \text{mol}$ $(34 \, g)$ of $NH_{3}$.
For $2.00 \times 10^{3} \, g$ of $N_{2}$,the required mass of $H_{2}$ is:
$\frac{6 \, g}{28 \, g} \times 2.00 \times 10^{3} \, g \approx 428.6 \, g$ of $H_{2}$.
Since we have $1.00 \times 10^{3} \, g$ of $H_{2}$ (which is more than $428.6 \, g$),$N_{2}$ is the limiting reagent.
The mass of $NH_{3}$ produced is determined by the limiting reagent $(N_{2})$:
$\text{Mass of } NH_{3} = \frac{34 \, g}{28 \, g} \times 2000 \, g \approx 2428.57 \, g$.
$(ii)$ Yes,$H_{2}$ will remain unreacted because it is in excess.
$(iii)$ The mass of $H_{2}$ remaining unreacted is:
$1000 \, g - 428.6 \, g = 571.4 \, g$.