Question

The number of moles of $KMn{O_4}$ required to oxidised $1$ mol of $Fe\left( {{C_2}{O_4}} \right)$ is;
A) $0.6$
B) $1.67$
C) $0.2$
D) $0.4$

Answer 1

We have to know that in a balanced chemical reaction, we predict the side product of the reaction. In some cases we are not able to attain balance by changing the mole. In that case we use the number of countable ions in the product and reactant side of the chemical reaction. In chemistry, redox reaction is one of the types of major reactions. In this redox reaction two methods are used to attain the balanced chemical reaction. There are ion-electron methods and the oxidation number method.
Moles are defined as the given mass of the molecule divided by the molecular mass of the molecule.
${\text{Moles}}\,{\text{ = }}\dfrac{{{\text{mass of the}}{\text{molecule}}}}{{{\text{molecular weight of the molecule}}}}$

Complete answer:
The chemical reaction for the oxidation of $Fe\left( {{C_2}{O_4}} \right)$ is given below,
$Fe\left( {{C_2}{O_4}} \right) \to F{e^{ + 3}} + 2C{O_2} + 3{e^ - }$
The individual reaction of each ion is given below,
$Fe\left( {{C_2}{O_4}} \right) \to F{e^{2 + }} + {\left( {{C_2}{O_4}} \right)^{2 - }}$
$F{e^{2 + }} \to F{e^{3 + }} + 1{e^ - }$
${\left( {{C_2}{O_4}} \right)^{2 - }} \to 2C{O_2} + 2{e^ - }$
According to the above discussion, there are three electrons released in the oxidation of $Fe\left( {{C_2}{O_4}} \right)$.
The chemical reaction for $KMn{O_4}$ to act as an oxidising agent in acidic medium is given below,
$Mn{O_4}^ - + 8{H^ + } + 5{e^ - } \to M{n^{2 + }} + 4{H_2}O$
We want to balance the number of electrons in both chemical reactions.
Hence, we increase the number of moles from one to five in the first chemical reaction and increase the number of moles from one to three in the second chemical reaction.
So, the chemical reaction is
$5F{e^{2 + }} + 5{\left( {{C_2}{O_4}} \right)^{2 - }} + 3Mn{O_4}^ - + 24{H^ + } \to 3M{n^{2 + }} + 12{H_2}O + 5F{e^{ + 3}} + 10C{O_2}$
From the above chemical reaction, we know $5$ moles of $Fe\left( {{C_2}{O_4}} \right)$ required $3$ moles of $KMn{O_4}$.
So, we calculate the number of moles of $KMn{O_4}$ required to oxidised $1$ mol of $Fe\left( {{C_2}{O_4}} \right)$ is given below,
$= \dfrac{3}{5} = 0.6$
According to the above discussion, we conclude the number of moles of $KMn{O_4}$ required to oxidised $1$ mol of $Fe\left( {{C_2}{O_4}} \right)$ is $0.6$.

Hence, option A is the correct answer.

Note:
As we know that in chemistry, equations play a major role. In the reaction three things are main. There are reactants, products and reaction conditions. The reactants are always on the left side of the equation. If more than one reactant in the equation means by using the plus sign. The products are always on the right side of the reaction. Something here also follows. More than one product by using plus sign. In between reactant products arrows are used to find the reaction direction in the equation.