Question

In the chemical reaction between stoichiometric quantities of \[KMn{O_4}\] and \[KI\] in weakly basic solution, what is the number of moles of \[{I_2}\] released for \[4\] moles of \[KMn{O_4}\] consumed?

Answer 1

Hint: The chemical equation for the given reaction can be written as follows:
\[KMn{O_4} + KI\xrightarrow{{{\text{Weak basic}}}}Mn{O_2} + {I_2}\]
To solve this question, we have to calculate the n factor.

Complete Step by Step Solution:
n factor or valency factor is a term used in redox reactions. It's a term that describes how many electrons per atom are gained or lost during a process.
In the given reaction, there is reduction of \[KMn{O_4}\] as follows:
\[KMn{O_4}\xrightarrow{{}}Mn{O_2}\]
Here, the oxidation state of \[Mn\] in \[KMn{O_4}\] is \[ + 7\] and the oxidation state of \[Mn\] in \[Mn{O_2}\] is \[ + 4\] . therefore, n factor is 3.
\[2{I^ - }\xrightarrow{{}}{I_2}\]
For \[{I_2}\] , the n factor is 2.
We know that, the equivalents will remain unchanged in the reaction.
So, we can say that,
\[ {\text{Equivalents of }}KMn{O_4} = {\text{Equivalents of }}{I_2} \\$
$ \Rightarrow n{\text{ factor}} \times {\text{moles of }}KMn{O_4} = n{\text{ factor}} \times {\text{moles of }}{{\text{I}}_{\text{2}}} \\$
$ \Rightarrow 3 \times 4 = 2 \times {\text{moles of }}{{\text{I}}_{\text{2}}} \\ $

Further solving this,
$ {\text{moles of }}{{\text{I}}_{\text{2}}} = \dfrac{{3 \times 4}}{2} \\$
$ \Rightarrow {\text{moles of }}{{\text{I}}_{\text{2}}} = 6 \\ $
As a result, moles of \[{I_2}\] is 6.

Additional Information: The normality of the solution is defined as the gram equivalent weight of a solute per litre of the solution. The mass of a substance is specified as the gram equivalent, but the mass of one equivalent substance is defined as the equivalent mass. Thus, equivalent mass can be calculated by dividing a substance's molar mass by its valence number/factor, whereas gram equivalent can be calculated by multiplying a substance's moles by its valence number.

Note: The type of reactant we use to calculate the n factor affects the valency factor computation. The n factor of a substance is equal to the product of the displaced mole and the charge of the product in non-redox reactions.