Question

50 kg of N2(g) and 10kg of $H_2\left(g\right)$ are mixed to produce $N{H}_3\left(g\right)$ . Calculate the amount of $N{H}_3$ formed. Identify the limiting reagent in the production of the $N{H}_3$.

Answer 1

As we know that,

$\text{no. of moles}=\frac{\text{Wt.}}{\text{mol. wt.}}$

Weight of $N_2=50kg$

Molecular weight of $N_2=28g$

No. of moles of $N_2=\frac{50\times {10}^3}{28}=17.86\times {10}^2\text{moles}$

Weight of $H_2=10kg$

Molecular weight of $N_2=2g$

No. of moles of $N_2=\frac{10\times {10}^3}{2}=5\times {10}^3\text{moles}$

$N_2+3H_2⟶2N{H}_3$

From the above reaction,

$1$ mole of $N_2$ react with $3$ moles of $H_2$.

No. of moles of $H_2$ required to react with $17.86\times {10}^2$

moles of $N_2=3\times 17.86\times {10}^2=5.36\times {10}^3\text{moles}$

But only $5\times {10}^3$ moles of $H_2$ are available.

Thus $H_2$ is the limiting reagent here.

Now, again from the above reaction,

Amount of ammonia formed when $3$ moles of $H_2$ react $=2\text{moles}$

Therefore,

Amount of ammonia formed when $5\times {10}^3$ moles of $H_2$

react $=\frac{2}{3}\times (5\times {10}^3)=3.33\times {10}^3\text{moles}$

Molecular weight of ammonia $=17g$

Weight of ammonia in $3.33\times {10}^3$ moles $=17\times (3.33\times {10}^3)=56.61kg$.

Hence the amount of $N{H}_3$ formed is $56.61kg$.