Question

50.0 kg of $N_2\left(g\right)$ and 10.0 kg of $H_2\left(g\right)$ are mixed to produce $N{H}_3\left(g\right).$ Calculate the mass of $N{H}_3$(g) formed:

• A. 40.0 kg
• B. 25.0 kg
• C. 56.6 kg
• D. 14.0 kg

Answer 1

The correct option is C 56.6 kg

$N_2+3H_2\to 2N{H}_3$
Moles of $N_2=\frac{\text{given mass}}{\text{molar mass}}=50\times \frac{1000}{28}=1.78\times {10}^3$ moles
Moles of $H_2=\frac{\text{given mass}}{\text{molar mass}}=10\times \frac{1000}{2}=5\times {10}^3$ moles
Finding the limiting reagent:
1 mole of $N_2$ reacts with 3 moles of $H_2$
$1.78\times {10}^3$ moles of $N_2$ will react with $3\times 1.78\times {10}^3=5.34\times {10}^3$ moles of $H_2$
Hence, the limiting reagent is hydrogen.
As per the reaction, 3 moles of $H_2$ forms 2 moles of $N{H}_3$
$5\times {10}^3$ moles of $H_2$ will form $\frac{2}{3}\times 5\times {10}^3=3.33\times {10}^3$ moles of $N{H}_3$
Mass of $N{H}_3$ formed $=3.33\times {10}^3\times 17=56.6$ kg