Question

In the following reaction:
$2{NH}_3⟶N_2+2{NH}_3$
Equivalent weight of ${NH}_3$ is $=\frac{Molecularweight}{y}$
What is the value of $y$?

Answer 1

In the following reaction:
$2{NH}_3⟶N_2+3H_2$
The equivalent weight of ${NH}_3$ is $=\frac{Molecularweight}{y}$
The formula for the equivalent mass is as given below.

Equivalent mass $=\frac{\text{Formula mass}}{\text{Total change in the oxidation number of an element per molecule that undergoes oxidation or reduction}}$

The oxidation number of $N$ changes from $-3$ to $0$.
The change in the oxidation number of $N$ per ammonia molecule is $3$.
Hence, the value of $y$ is 3.