Question

Hydrazine $\left(N_2{H}_4\right)$ is a rocket fuel which can be produced according to the following reaction:

$ClN{H}_2+2N{H}_3\to N_2{H}_4+N{H}_{4}Cl$

When $1000$ g of $ClN{H}_2$ is reacted with excess of $N{H}_3$, $473$ g of $N_2{H}_4$ is produced. What is the percentage yield of the reaction?

• A. $76.12$
• B. $67.21$
• C. $26.17$
• D. $16.72$

Answer 1

The correct option is A $76.12$

$1$ mole of $ClN{H}_2$ reacts to form $1$ mole of $N_2{H}_4$.

The molar mass of $ClN{H}_2$ is $35.5+14+2=51.5$ g/mol
The molar mass of $N_2{H}_4$ is $2\left(14\right)+4=32$ g/mol

Thus, $51.5$ g of $ClN{H}_2$ reacts to form $32$ g of $N_2{H}_4$.

Thus, $1000$ g of $ClN{H}_2$ reacts to form $\frac{32\times 1000}{51.5}=621.36$ g of $N_2{H}_4$.

However, only $473$ g of $N_2{H}_4$ are obtained.

Hence, the percentage yield of the reaction is $\frac{473}{621.36}\times 100=76.12\%$.

Therefore, option A is correct.