50 kg of  $N_2\left(g\right)$ and 10 kg of  $H_2$ (g) are mixed to produce $N{H}_3\left(g\right)$ . The correct statements about the reaction is ?

$\begin{array}{l}{N}_{2\left(g\right)}+3H_{2\left(g\right)}\to 2N{H}_3\left(g\right)\\ moles=\frac{50,000}{28}\frac{10,000}{2}\\ 1785moles5,000moles\\ \frac{1785}{1}\frac{5,000}{3}\\ =17851666.6\left(lesser\right)\\ \left(excessreagent\right)\left(\therefore Limitingreagent\right)\end{array}$

Amount of product formed is calculated as per limiting reagent

$\begin{array}{l}3moles{H}_2\to 2molesN{H}_3\\ 5,000moles\to ?\\ =\frac{5000\times 2}{3}=3333.33moles\\ =3333.3\times 17gr\\ =56.666gr\\ =56.6kg\end{array}$