Question

$N_2+3H_2⟶2N{H}_3$
Here, the equivalent mass of $N_2=\frac{28}{x}$. The value of $x$ is :

Answer 1

The balanced reaction is as follows:
$N_2+3H_2⟶2N{H}_3$
0 0 -3 +1
As we know,
Equivalent weight $=\frac{molarweight}{n-factor}$
So,
$E_{N_2}=\frac{28}{6}=4.67$
So, value of $x$ is $6$.