Question

$N_2+3H_2\to 2N{H}_3$
Molecular weight of $N{H}_3$ and $N_2$ are $x_1$ and $x_2$ respectively. Their equivalent weights are $y_1$ and $y_2$ respectively. Then $(y_1-y_2)$ is :

• A. $\left(\frac{2x_1-x_2}{6}\right)$
• B. $(x_1-x_2)$
• C. $(3x_1-x_2)$
• D. $(x_1-3x_2)$

Answer 1

Answer: (A)

$\left(\frac{2x_1-x_2}{6}\right)$

$N_2\equiv 2N{H}_3\equiv 3H_2\equiv 6H$
Ew of $N_2=\frac{x_2}{6}=y_2$
Ew of $N{H}_3=\frac{2x_1}{6}=y_1$
$\therefore y_1-y_2=(\frac{2x_1}{6}-\frac{x_2}{6})=\left(\frac{2x_1-x_2}{6}\right)$