Question

How can I balance this equation?
${\text{KCl}}{{\text{O}}_{\text{3}}} \to {\text{KCl + }}{{\text{O}}_{\text{2}}}$

Answer 1

For balancing any chemical equation, we have to maintain equal molecularity of each atom present on the reactant side as well as on the product side of the given chemical reaction.

Complete step by step solution:
Given that, Potassium chlorate produces potassium chloride and hydrogen gas and chemical reaction for this is shown as follow:
${\text{KCl}}{{\text{O}}_{\text{3}}} \to {\text{KCl + }}{{\text{O}}_{\text{2}}}$
For balancing the above given equation we have to consider following points in mind:
- First we calculate the number of each atom on the left hand side as well as on the right hand side of the given chemical equation.
- So on the left hand side or reactant side and on the right hand side or product side of the chemical reaction only one Potassium (${\text{K}}$) atom is present, which is correct.
- Similarly on the left hand side or reactant side and on the right hand side or product side of the chemical reaction only one atom of chlorine (${\text{Cl}}$) is present, which is also correct.
- But on the left side three oxygen atoms (${\text{O}}$) and on the right side two oxygen atoms (${\text{O}}$) are present which is not correct.
So, we will balance the above given reaction by the following manner so that each atom will have the same molecularity on the reactant side as well as on the product of the reaction. And balanced equation is shown as:
${\text{2KCl}}{{\text{O}}_{\text{3}}} \to 2{\text{KCl + 3}}{{\text{O}}_{\text{2}}}$
- In the above balanced equation two potassium (${\text{K}}$) atom, two chlorine atoms (${\text{Cl}}$) and six oxygen (${\text{O}}$) atoms are present on the left hand side as well as on the right hand side of the chemical reaction also.

Note: Here some of you may think order of reaction and molecularity of reaction are the same things but that will be true only for the elementary (single step reaction) reactions only.