$50.0 \, kg$ of $N_2 \, (g)$ and $10.0 \, kg$ of $H_2 \, (g)$ are mixed to produce $NH_3 \, (g)$. Calculate the amount of $NH_3 \, (g)$ formed. Identify the limiting reagent in the production of $NH_3$ in this situation.

(N/A) The balanced chemical equation for the reaction is:
$N_2 \, (g) + 3H_2 \, (g) \rightleftharpoons 2NH_3 \, (g)$
Calculation of moles:
Number of moles of $N_2 = 50.0 \, kg \times \frac{1000 \, g}{1 \, kg} \times \frac{1 \, mol}{28.0 \, g} = 1786 \, mol$
Number of moles of $H_2 = 10.0 \, kg \times \frac{1000 \, g}{1 \, kg} \times \frac{1 \, mol}{2.016 \, g} = 4960 \, mol$
According to the stoichiometry,$1 \, mol$ of $N_2$ requires $3 \, mol$ of $H_2$.
For $1786 \, mol$ of $N_2$,the required $H_2 = 1786 \times 3 = 5358 \, mol$.
Since we have only $4960 \, mol$ of $H_2$,$H_2$ is the limiting reagent.
Calculation of $NH_3$ formed:
$3 \, mol$ of $H_2$ produces $2 \, mol$ of $NH_3$.
$4960 \, mol$ of $H_2$ produces $\frac{2}{3} \times 4960 = 3306.67 \, mol$ of $NH_3$.
Mass of $NH_3 = 3306.67 \, mol \times 17.03 \, g/mol \approx 56300 \, g = 56.3 \, kg$.