Question

Balance the following equation:
$KCl{O_3} + {H_2}S{O_4} \to KHS{O_4} + HCl{O_4} + Cl{O_2}$

$A.\;3KCl{O_3} + 3{H_2}S{O_4} \to 3KHS{O_4} + HCl{O_4} + 2Cl{O_2} + {H_2}O$
$B.\;KCl{O_3} + 5{H_2}S{O_4} \to 3KHS{O_4} + 3HCl{O_4} + 2Cl{O_2} + 3{H_2}O$
$C.\;3KCl{O_3} + 4{H_2}S{O_4} \to 3KHS{O_4} + 2HCl{O_4} + 2Cl{O_2} + {H_2}O$
$D.\;None\;of\;these$

Answer 1

Hint: - For answering these types of questions we must count the number of atoms of each element in the reactants and in the products side.

Complete step-by-step answer:
The unbalanced equation given to us is
$KCl{O_3} + {H_2}S{O_4} \to KHS{O_4} + HCl{O_4} + Cl{O_2}$
Initially we will balance all the atoms excluding $H$and $O$
$ \Rightarrow 2KCl{O_3} + 2{H_2}S{O_4} \to 2KHS{O_4} + HCl{O_4} + Cl{O_2}$
The number of oxidations of $Cl$ increases from +5 to +7 and the increase in the number of atoms is 2, the number of oxidations of $Cl$ increases from +5 to 4 and the increase in the number of atoms is 1 and the increase in the number of oxidations is offset by a decrease in the number of oxidations by multiplying $KCl{O_3}$ and $Cl{O_2}$ with 2.
Balancing K in the equation: -
$ \Rightarrow 3KCl{O_3} + 2{H_2}S{O_4} \to 2KHS{O_4} + HCl{O_4} + 2Cl{O_2}$
Now we can see that $O$ atoms are balanced by adding 1 water molecule on Right hand side.
$ \Rightarrow 3KCl{O_3} + 3{H_2}S{O_4} \to 3KHS{O_4} + HCl{O_4} + 2Cl{O_2} + {H_2}O$
After this we can observe that all the molecules are balanced. And our final equation comes out to be: -
$ \Rightarrow 3KCl{O_3} + 3{H_2}S{O_4} \to 3KHS{O_4} + HCl{O_4} + 2Cl{O_2} + {H_2}O$
Hence option A is the correct answer and the balanced equation of the given question is
$3KCl{O_3} + 3{H_2}S{O_4} \to 3KHS{O_4} + HCl{O_4} + 2Cl{O_2} + {H_2}O$

Note: For any chemical equation the reactant mass must be equal to the product mass. We have to count the number of atoms in the reactants of each element, and the number of atoms in the products of each element. If the equation isn't balanced, change the coefficients of the molecules on each side of the equation balance until the number of atoms in each element.