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

Apr 22, 2018

Answer:

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 $K C l {O}_{3}$ and is reduced to $C l -$ which has an oxidation state of $- 1$.
Oxygen oxidation state is $- 2$ in $K C l {O}_{3}$ and is oxidised to ${O}_{2}$ which has an oxidation state of $0$.
Write Half Reactions.
$C {l}^{5 +} + 6 e = C {l}^{-}$
$2 {O}^{2 -} = {O}_{2} + 4 e$
Balance electrons in accordance to conservation of charges.
$4 C {l}^{5 +} + 24 e = 4 C {l}^{-}$
$12 {O}^{2 -} = 6 {O}_{2} + 24 e$

Thus interpreting this means that there is $4 K C l {O}_{3}$, $4 K C l$, $6 {O}_{2}$. This balances both atoms and charges making it fully balanced.
Thus simplify the equation by a factor of 2,
$2 K C l {O}_{3} = 2 K C l + 3 O 2$