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 $2.00\times {10}^{2}g$ dinitrogen react with $1.00\times {10}^{3}g$ of dihydrogen
(iii) Will any of the two reactants remain unreacted?
(iii) If yes, which one and what would be its mass?

Answer 1

$\underset{\underset{=28.0g}{1mol}}{Na\left(g\right)}+\underset{\underset{=6.0g}{3mol}}{3H_{2\left(g\right)}}\to \underset{\underset{=34.0g}{2mol}}{2N{H}_{3\left(g\right)}}$

(i) $28.0g$ of $N_2$ require $6.0g$ of $H_2$ to produce

$=34.0g$ of $N{H}_3$

$2.00\times {10}^{3}g$ of $N_2$ will produce

$=\frac{34}{28}\times 2.00\times {10}^{3}g$ of $N{H}_3$

$=2.43\times {10}^{3}g$ of $N{H}_3$

$=2430gN{H}_3$

(ii) Yes, dihydrogen will remain unreacted to some extent

(iii) Amount of hydrogen that remains unreacted.

$28.0g$ of $N_2$ require $6.0g$ of $H_2$

$2.00g\times {10}^{3}g$ of $N_2$ will require

$=\frac{6.0}{28.0}\times 2.00\times {10}^3$ of $H_2=428.5g$ of $H_2$

$\therefore$ Amount of hydrogen that remains unreacted

$=[1.00\times {10}^3-428.5{]}_g$

$=571.5g$.