Question

Hydrazine $\left(N_2{H}_4\right)$ can be produced according to the following reaction:
$ClN{H}_2+2N{H}_3\to N_2{H}_4+N{H}_{4}Cl$. 1 kg of $ClN{H}_2$ is reacted with excess $N{H}_3$, find the actual number of molecules of $N_2{H}_4$ produced if the percentage yield of the given reaction is 76%.

• A. $15N_A$
• B. $13N_A$
• C. $17N_A$
• D. $3.2N_A$

Answer 1

The correct option is A $15N_A$

$ClN{H}_2+2N{H}_3\to N_2{H}_4+N{H}_{4}Cl$
It is given that 1 kg of $ClN{H}_2$ reacts with excess $N{H}_3$. Percentage yield of reaction is given as 76%, we need to find number of molecules of $N_2{H}_4$
Moles of $ClN{H}_2=\frac{\text{Mass}}{\text{Molar mass}}=\frac{1000\text{g}}{51.5{\text{g mol}}^{-1}}=19.4\text{mol}$
From the balance reaction, it is clear that 1 mole of $ClN{H}_2$ produces 1 mole of $N_2{H}_4$. Hence, 19.4 mol of $ClN{H}_2$ produces 19.4 mol of $N_2{H}_4$.
From the percentage yield, the given actual moles of $N_2{H}_4$ will be
Actual moles of $N_2{H}_4$ = $19.4\times 0.76=14.74\text{mol}$
$\therefore \text{Number of molecules of}{N}_2{H}_4=14.74N_A$