Question

$N_2+3H_2⟶2N{H}_3$
The equivalent weight of $N{H}_3=\frac{17}{x}$. The value of $x$ is:

Answer 1

Reaction:
$N_2+3H_2⟶2N{H}_3$
0 0 -3 +1
As we know,
Equivalent wt $=\frac{molarwt.}{n-factor}$
So,
$E_{N{H}_3}=\frac{17}{3}=5.67$
So, value of $x$ is $3$.