Question

In the reaction$Zn+HN{{O}_{3}}\to Zn{{(N{{O}_{3}})}_{2}}+NO+{{H}_{2}}O$, the equivalent weight of $HN{{O}_{3}}$ is _____.
(A) M
(B) $4M/3$
(C) $8M/3$
(D) $2M/3$

Answer 1

Hint: Equivalent weight also known as gram equivalent weight is the mass of one equivalent which means the mass of a given substance which will relocate or combines with a fixed quantity of the other compound and equivalent weight can also be expressed as the mass which combines or displaces 1.008 grams of hydrogen or about 8.0 g of oxygen or 35.5 g of chlorine.

Complete step by step solution:
To calculate the equivalent weight first we will calculate the n-factor for$HN{{O}_{3}}$ and then it will be calculated by the ratio of molecular weight to n- factor.
Given reaction is:
$Zn+HN{{O}_{3}}\to Zn{{(N{{O}_{3}})}_{2}}+NO+{{H}_{2}}O$
The balanced chemical reaction is:
$2Zn+8HN{{O}_{3}}\to 3Zn{{(N{{O}_{3}})}_{2}}+2NO+4{{H}_{2}}O$
Now we have to calculate the oxidation state of nitrogen in$HN{{O}_{3}}$ .
Oxidation state of H= 1
Oxidation state of O=-2
Oxidation state of N= $1+N+3X(-2)=0$
Oxidation state of N=+5
Now we have to calculate the oxidation state of nitrogen in NO
Oxidation state of O=-2
Oxidation state of N= $N-2=0$
Oxidation state of N= +2
There is change in the oxidation state of nitrogen from +5 to +2
And from the balanced reaction we can say that, 8 $HN{{O}_{3}}$ reacts to form 2NO.
So, the n factor for 1 mole of $HN{{O}_{3}}$= $\dfrac{2}{3}X3=\dfrac{3}{4}$
So the equivalent weight of $HN{{O}_{3}}$ is calculated by the ratio of molecular weight to n- factor
Equivalent weight of $HN{{O}_{3}}$= $\dfrac{\text{molecular weight}}{n-factor}$
=\[\dfrac{M}{(\dfrac{3}{4})}=\dfrac{4M}{3}\]

Hence the correct option is option (B).

Note: In a type of redox reaction where the same compound undergoes oxidation and reduction simultaneously, the calculation of n-factor is required as it plays a crucial role in determining the equivalent weight.