Dinitrogen and dihydrogen react with each other to produce ammonia according to the following chemical equation: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$
$(i)$ 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) The balanced chemical equation is: $N_{2(g)} + 3H_{2(g)} \to 2NH_{3(g)}$
From the stoichiometry: $28 \ g$ of $N_2$ reacts with $6 \ g$ of $H_2$ to produce $34 \ g$ of $NH_3$.
Given: Mass of $N_2 = 2000 \ g$,Mass of $H_2 = 1000 \ g$.
Step $1$: Identify the limiting reagent.
For $2000 \ g$ of $N_2$,the required $H_2 = (2000 \ g \ N_2 \times 6 \ g \ H_2) / 28 \ g \ N_2 = 428.57 \ g \ H_2$.
Since we have $1000 \ g$ of $H_2$ (which is more than $428.57 \ g$),$N_2$ is the limiting reagent.
$(i)$ Mass of $NH_3$ produced = $(2000 \ g \ N_2 \times 34 \ g \ NH_3) / 28 \ g \ N_2 = 2428.57 \ g \ NH_3$.
$(ii)$ Yes,$H_2$ will remain unreacted.
$(iii)$ Mass of unreacted $H_2 = 1000 \ g - 428.57 \ g = 571.43 \ g$.