Question

$a{\text{KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}b{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + c{\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}$ In this unbalanced stoichiometric equation the value of $a,b{\text{ and }}c$respectively are

Answer 1

Chemical equation is the symbolic representation of a chemical reaction. The balanced chemical equation is an expression of the law of conservation of mass. The law states that the total amount of matter in the universe is unaltered whatever changes take place in its distribution.

Complete step-by-step answer:
A chemical reaction is the conversion of one or more reactants into one or more products and a chemical equation is the representation of chemical formulas of reactants and products.

The chemical equation given to us is
$a{\text{KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}b{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + c{\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}$
Here we have already given both reactants and products. We have to determine the values of $a,b{\text{ and }}c$. For that, we have to balance the chemical reaction using the law of conservation of mass.
The Law of conservation of mass states that the mass of product and mass of reactant is always equal. In another word, it states that there is no loss or gain of atoms.
To balance the reaction we have to add an appropriate coefficient to the chemical species (Reactant /product) without changing the chemical formulas or adding new species.
First, we will balance the ${\text{K}}$ atom on both sides of the equations.
$a{\text{KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}b{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + c{\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}$

Here we can see that there are 2 atoms of ${\text{K}}$ on the product side and one atom of ${\text{K}}$on the reactant side so to balance the ${\text{K}}$ atoms we will add coefficient 2 to ${\text{KMn}}{{\text{O}}_{{\text{4 }}}}$.
${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}$

Now, we will balance the ${\text{Mn}}$ atoms by adding coefficient 2 to ${\text{MnS}}{{\text{O}}_{\text{4}}}$.

${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{O }}$

In this reaction ${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}$ corresponds to 8 ‘${\text{O}}$’ atoms so to balance it we need to add coefficient 8 to${{\text{H}}_{\text{2}}}{\text{O }}$.
${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{O }}$

Now, to balance the ‘${\text{H}}$’ atoms we will add coefficient 8 to ${{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$.
${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + {\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2MnS}}{{\text{O}}_{\text{4}}}{\text{ + F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{O }}$
Now, to balance the ${\text{Fe}}$ and ${\text{S}}{{\text{O}}_{\text{4}}}$ we have to add coefficient 10 to ${\text{FeS}}{{\text{O}}_{\text{4}}}$ and coefficient 5 to ${\text{F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}$
${\text{2KMn}}{{\text{O}}_{{\text{4 }}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} + 10{\text{FeS}}{{\text{O}}_{\text{4}}} \to {\text{ }}{{\text{K}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + 2MnS}}{{\text{O}}_{\text{4}}}{\text{ + 5F}}{{\text{e}}_{\text{2}}}{{\text{(S}}{{\text{O}}_{\text{4}}}{\text{)}}_{\text{3}}}{\text{ + 8}}{{\text{H}}_{\text{2}}}{\text{O }}$

This is a balanced chemical equation. So, from this balance reaction, we can say that the values of $a,b{\text{ and }}c$ are 2,8 and 10.

Hence, the correct answer is an option (A) 2,8,10

Note: Stoichiometry is a measure of elements. It is based on the law of conservation of mass. So the concept of stoichiometry is used to determine the amount of products or reactants, using a mole ratio of a balanced chemical reaction. While balancing any chemical reaction we can only change the coefficients of reacting species. We cannot change the subscripts of the chemical formula as it will cause a change in the chemical identity of reacting species.