Question

Potassium chlorate $\left( {{\text{KCl}}{{\text{O}}_3}} \right)$ on heating gives potassium chloride and oxygen. Write a balance for this reaction and indicate the evolution of gas.

Answer 1

First count the number of atoms of each type of element on the reactants side as well as on the products side. Then balance the number of each type of element by using stoichiometric coefficients.

Complete answer:
Potassium chlorate is represented by the chemical formula ${\text{KCl}}{{\text{O}}_3}$ .
The thermal decomposition of potassium chlorate forms potassium chloride and oxygen. Write a balanced for this reaction
${\text{KCl}}{{\text{O}}_3}\left( s \right){\text{ + heat }} \to {\text{ KCl}}\left( s \right){\text{ + }}{{\text{O}}_2}\left( g \right) \uparrow$
The evolution of oxygen gas is represented by an upward arrow.
In the above equation, potassium atoms are already balanced. Similarly, chlorine atoms are already balanced. But oxygen atoms need to be balanced.
On the reactants side, 3 oxygen atoms are present. On the products side, two oxygen atoms are present. To balance the oxygen atoms, multiply the oxygen molecule on the products side with 3/2.

${\text{KCl}}{{\text{O}}_3}\left( s \right){\text{ + heat }} \to {\text{ KCl}}\left( s \right){\text{ + }}\dfrac{3}{2}{{\text{O}}_2}\left( g \right) \uparrow$

This is the balanced chemical equation.
The above balanced chemical equation contains fraction. But it is desirable that the coefficients of all the species should be whole numbers. For this purpose, eliminate the fraction by multiplying the whole equation with 2.
${\text{2KCl}}{{\text{O}}_3}\left( s \right){\text{ + heat }} \to {\text{ 2KCl}}\left( s \right){\text{ + }}3{{\text{O}}_2}\left( g \right) \uparrow$

This is the balanced chemical equation for the thermal decomposition of potassium chlorate to form potassium chloride and oxygen gas.

Note: Thermal decomposition of potassium chlorate is carried out in presence of manganese dioxide $\left( {{\text{Mn}}{{\text{O}}_2}} \right)$ catalyst. A catalyst provides an alternate pathway of lower activation energy, thereby, the catalyst increases the rate of the reaction.