Question

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

$N_2\left(g\right)+H_2\left(g\right)\to 2{NH}_3\left(g\right)$

(i) Calculate the mass of ammonia produced if $2.00\times {10}^3$ g 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?

Answer 1

The balanced chemical equation is as follows:

$N_2+3H_2\to 2N{H}_3$
The molar masses of nitrogen, hydrogen and ammonia are $28$ g/mol, $2$ g/mol and $17$ g/mol respectively.

$28$ g of nitrogen reacts with $6$ g of hydrogen to form $34$ g of ammonia.

$2.00\times {10}^3$ g of nitrogen will react with $\frac{6}{28}\times 2.00\times {10}^3=428.6$ g hydrogen.

However, $1.00\times {10}^3$g of dihydrogen are present.

Hence, nitrogen is the limiting reagent.
(i) Mass of ammonia produced $=\frac{34}{28}\times 2\times {10}^3=2428.57$ g
(ii) Hydrogen is the excess reagent. Hence, it will remain unreacted.
(iii) Mass of dihydrogen left unreacted $=1.00\times {10}^3-428.6=571.4$ g