50.0kg of $N_{2\left(g\right)}$  and 10.0kg of $H_{2\left(g\right)}$  are mixed to produce $N{H}_{3\left(g\right)}$ . 

Identify the limiting reagent and find the amount of $N{H}_{3\left(g\right)}$  formed

$N_{2\left(g\right)}+3H_{2\left(g\right)}\rightleftharpoons 2N{H}_{3\left(g\right)}$ 

$\left(2\times 14\right)g$   6g                 $2\left(14+3\right)g$

28 kg         6 kg             34 kg

According to the balanced chemical reaction, 28kg of $N_2$  reacts with 6kg of $H_2$  to give 34 kg of $N{H}_3$ .

Thus, 6 kg of $H_2$  requires 28 kg of $N_2$

10kg of $H_2$  requires $\frac{28\times 10}{6}=46.6kgof{N}_2$

Hence, it is the limiting reagent, since it is completely consumed in the reaction, and $H_2$  will decide the amount of product formed.

Thus 6 kg of $H_2=34kgofN{H}_3$

10 kg of $H_2=\frac{34\times 10}{6}=56.6kgofN{H}_3$