Question

How do you balance `KClO_3(s) -> KCl(s) + O_2(g)`?

Answer 1

Using oxidation states.

Explanation:

We have to realize that this is a redox reaction.
Chlorine is being reduced while oxygen is being oxidized.
Chlorine oxidation state is `+5` in `KClO_3` and is reduced to `Cl-` which has an oxidation state of `-1`.
Oxygen oxidation state is `-2` in `KClO_3` and is oxidised to `O_2` which has an oxidation state of `0`.
Write Half Reactions.
`Cl^(5+) + 6e = Cl^-`
`2O^(2-) = O_2 + 4e`
Balance electrons in accordance to conservation of charges.
`4Cl^(5+) + 24e = 4Cl^-`
`12O^(2-) = 6O_2 + 24e`

Thus interpreting this means that there is `4KClO_3`, `4KCl`, `6O_2`. This balances both atoms and charges making it fully balanced.
Thus simplify the equation by a factor of 2,
`2KClO_3 = 2KCl + 3O2`