How do you balance ${\mathrm{KClO}}_3\to \mathrm{KCl}+{\mathrm{O}}_2$?

Step 1: Note down the unbalanced reaction.

Write the unbalanced reaction as it is.

$\underset{\mathrm{Potassium}\mathrm{chlorate}}{{\mathrm{KClO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloirde}}{\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{{\mathrm{O}}_2\left(\mathrm{g}\right)}$

Step 2: Balance all the elements except $\mathrm{O}$.

To balance $\mathrm{K}$: On both the right and left sides of the reaction, there is only one $\mathrm{K}$ is present, therefore there is no need to balance it.

To balance $\mathrm{Cl}$: Chlorine has the same number on both sides of the reaction, hence, there is no need to balance it.

Step 3: Balance $\mathrm{O}$

The number of Oxygen is not the same on both sides of the reaction, on the left side, there are three Oxygen atoms while only two are on the right-hand side. to balance the number of Oxygen on both sides, multiply ${\mathrm{KClO}}_3$ by $2$ and multiply ${\mathrm{O}}_2$ by $3$.

Hence, by multiplying chemical compounds with suitable coefficients, we get the balanced equation as $\underset{\mathrm{Potassium}\mathrm{chlorate}}{2{\mathrm{KClO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloirde}}{2\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$

Therefore, the balanced chemical reaction is$\underset{\mathrm{Potassium}\mathrm{chlorate}}{2{\mathrm{KClO}}_3\left(\mathrm{s}\right)}\to \underset{\mathrm{Potassium}\mathrm{chloirde}}{2\mathrm{KCl}\left(\mathrm{s}\right)}+\underset{\mathrm{Oxygen}}{3{\mathrm{O}}_2\left(\mathrm{g}\right)}$.