Question

Hydrazine reacts with $KI{{O}_{3}}$in the presence of $HCl$ as:
\[{{N}_{2}}{{H}_{4}}+IO_{3}^{-}+2{{H}^{-}}+C{{l}^{-}}\to ICl+{{N}_{2}}^{+}+3{{H}_{2}}O\]
The equivalent masses of ${{N}_{2}}{{H}_{4}}$and $KI{{O}_{3}}$respectively are: $[KI{{O}_{3}}=214]$
A. $8 and 53.5$
B. $16 and 53.5$
C. $8 and 35.6$
D. $8 and 87$

Answer 1

Hint: For finding the equivalent weight in the given question, first of all separate the equations into oxidation and reduce half reactions. Then, you will get the n-factors of ${{N}_{2}}{{H}_{4}}$and $KI{{O}_{3}}$from the reaction. And then, by applying the direct formula for equivalent mass you will get your answer.


Complete step by step solution:
Given that,
Hydrazine with chemical formula ${{N}_{2}}{{H}_{4}}$is reacting with $KI{{O}_{3}}$ in the presence of $HCl$ giving the following chemical equation:
\[{{N}_{2}}{{H}_{4}}+IO_{3}^{-}+2{{H}^{-}}+C{{l}^{-}}\to ICl+{{N}_{2}}^{+}+3{{H}_{2}}O\]
And, we have to find out the equivalent mass or weight of ${{N}_{2}}{{H}_{4}}$ and $KI{{O}_{3}}$.
So, for this we have to find the n-factors of ${{N}_{2}}{{H}_{4}}$and $KI{{O}_{3}}$.
So, divide the chemical reaction into two parts i.e. oxidation and reduction half reaction and we can get the n-factors.
Thus,
Oxidation and reduction half reactions are:
${{N}_{2}}{{H}_{4}}\to {{N}_{2}}+4{{H}^{+}}+4{{e}^{-}}$
$I{{O}_{3}}^{-}+6{{H}^{+}}+4{{e}^{-}}+C{{l}^{-}}\to ICl+3{{H}_{2}}O$
So here, we can see that,
Change in oxidation from ${{N}_{2}}{{H}_{4}}$to ${{N}_{2}}$is 4.
Hence, the n-factor of ${{N}_{2}}{{H}_{4}}$will be 4.
And, the molecular mass of ${{N}_{2}}{{H}_{4}}$is \[2\times 14+2\times 4=32g\].
Similarly,
Change in oxidation state from \[I{{O}_{3}}^{-}\]to \[ICl\]is 4.
Thus, the n-factor of $KI{{O}_{3}}$will be 4.
And, the molecular mass of $KI{{O}_{3}}$is \[39\times 1+127\times 1+3\times 16=214g\].
Now we can get our equivalent weights of the following compounds.
Using the formula,
$Equivalent Weight=\dfrac{Molecular Mass}{Number Of Electrons Gained Or Lost}$

So, the equivalent weight of ${{N}_{2}}{{H}_{4}}$using the above formula will be,
\[Equivalent Weight=\dfrac{32}{4}=8\]
Similarly, the equivalent weight of $KI{{O}_{3}}$using the same formula will be,
\[Equivalent Weight=\dfrac{214}{4}=53.5\]
Thus, the equivalent weight of ${{N}_{2}}{{H}_{4}}$is 8 and that of $KI{{O}_{3}}$is 53.5

Hence, the correct option is A.

Note: To calculate equivalent mass, simply divide the molar mass of the substance given by the number of electrons which is gained or lost during the it’s decomposition. It would be easy to know the n-factor, if you will divide the reaction into oxidation and reduce half reaction.