$50.0 \ kg$ 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 $NH_{3(g)}$ in this situation.

(N/A) The balanced chemical equation for the reaction is: $N_{2(g)} + 3H_{2(g)} \rightarrow 2NH_{3(g)}$
Number of moles of $N_{2} = 50.0 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol}{28.02 \ g} = 1784.4 \ mol$
Number of moles of $H_{2} = 10.0 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol}{2.016 \ g} = 4960.3 \ mol$
According to the stoichiometry,$1 \ mol$ of $N_{2}$ requires $3 \ mol$ of $H_{2}$.
Therefore,$1784.4 \ mol$ of $N_{2}$ requires $1784.4 \times 3 = 5353.2 \ mol$ of $H_{2}$.
Since we have only $4960.3 \ mol$ of $H_{2}$,$H_{2}$ is the limiting reagent.
The amount of $NH_{3}$ produced depends on the limiting reagent $(H_{2})$:
$3 \ mol$ of $H_{2}$ produces $2 \ mol$ of $NH_{3}$.
$4960.3 \ mol$ of $H_{2}$ produces $\frac{2}{3} \times 4960.3 = 3306.9 \ mol$ of $NH_{3}$.
Mass of $NH_{3} = 3306.9 \ mol \times 17.03 \ g/mol = 56316.5 \ g \approx 56.3 \ kg$.