Question

$50.0Kg$ of $N_2\left(g\right)$ and $10.0kg$ of $H_2\left(g\right)$ are mixed to produce ${NH}_3\left(g\right)$. Calculate the moles of ${NH}_3\left(g\right)$ formed. Identify the limiting reagent in the production of ${NH}_3$ in this situation.

• A. $H_2$
• B. $N{H}_3$
• C. $N_2$
• D. Cant be predicted

Answer 1

The correct option is A $H_2$

$N_2+2H_2\rightleftharpoons 2N{H}_3$

$50kg$ $10kg$

$No.ofmole=\frac{mass}{molarmass}n=\frac{x}{M.M}$

For $N_{2}n=\frac{50\times 1000}{28}=1.785\times {10}^3$

For $H_{2}n=\frac{10\times 1000}{2}=5\times {10}^3$

The ratio $\frac{n}{S.C.}=\frac{mole}{Stiochiometriccoeff}\Rightarrow$lowest$\Rightarrow$Limiting Reagent

For $N_2\frac{n}{S.C.}=\frac{1.785\times {10}^3}{1}=1.785\times {10}^3$

For $H_2\frac{n}{S.C.}=\frac{5\times {10}^3}{3}=1.66\times {10}^3$

Therefore $H_2$ is the limiting reagent

$\frac{n_{H_2}}{{SC}_{H_2}}=\frac{n_{N{H}_3}}{{SC}_{N{H}_3}}\frac{5\times {10}^3}{3}=\frac{n_{N{H}_3}}{2}\frac{2\times 5\times {10}^3}{3}=n_{N{H}_3}\frac{10\times {10}^3}{3}=n_{N{H}_3}\frac{{10}^4}{3}=n_{N{H}_3}=3333.33$

The moles of $N{H}_3\left(g\right)$ formed is $3333.33$