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?
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 $=1.00 \times 10^{3} \mathrm{~g}-428.6 \mathrm{~g}$
$=571.4 \mathrm{~g}$
(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 $=1.00 \times 10^{3} \mathrm{~g}-428.6 \mathrm{~g}$
$=571.4 \mathrm{~g}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.