$50.0$ kg of $N_2$ gas and $10.0$ kg of $H_2$ gas are mixed to produce $N{H}_3$. The limiting reagent and amount of $N{H}_3$ formed are:

The correct option is A $H_2$, $56.7$ kg
$N_2+3H_2\to 2N{H}_3$

$28$ kg of $N_2$ combines with $6$ kg of $H_2$ to form $34$ kg of $N{H}_3$.

$\therefore$ $10$ kg of $H_2$ (limiting reagent) will consume $46.7$ kg of $N_2$ to form $56.7$ kg of $N{H}_3$.

After entire $10$ kg of $H_2$ is consumed, the reaction will stop and $3.3$ kg of $N_2$ will be left.

Option A is correct.