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

1 Answer
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 #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#