What is the limiting reagent? $50kg{N}_2$ and $10kg{H}_2$ are mixed to produce $N{H}_3$. Calculate the ammonia gas formed. Also, identify the limiting reagent.

Solution:-

Limiting reagent $\to$ The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete.

Given:-

Weight of $N_2=50kg=5\times {10}^{4}g$

Weight of $H_2=10kg={10}^{4}g$

Molecular weight of $N_2=28g$

Molecular weight of $H_2=2g$

As we know that,

No. of moles $=\frac{\text{Weight}}{\text{Mol. wt.}}$

Therefore,

No. of moles of $N_2=\frac{5\times {10}^4}{28}=1.786\times {10}^3\text{moles}$

No. of moles of $H_2=\frac{{10}^4}{2}=5\times {10}^3\text{moles}$

Now,

$N_2+3H_2⟶2N{H}_3$

From the above reaction,

$1$ mole of $N_2$ reacts with $3$ moles of $H_2$ to produce $2$ mole of ammonia.

Therefore,

$1.786\times {10}^3$ moles of $N_2$ will react with $5.36\times {10}^3$ moles of $H_2$ to produce ammonia.

But given amount of $H_2$ is $5$ moles.

Thus $H_2$ is limiting reagent.

Therefore,

Amount of ammonia produced by $3$ moles of $H_2=2\text{moles}$

$\therefore$ Amount of ammonia produced by $5\times {10}^3$ moles of $H_2=\frac{2}{3}\times 5\times {10}^3=3.33\times {10}^3\text{moles}$

Hence $3.33$ moles of ammonia gas is formed.