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

Question

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

Owen Ambrose · Accepted Answer

Using oxidation states.

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#

Serenity Allen · Answer

undefined To balance the chemical equation $ KClO_3(s) \rightarrow KCl(s) + O_2(g) $, you would need to adjust the coefficients of the reactants and products so that the number of atoms of each element is the same on both sides of the equation. In this case, you would balance it as follows:

$$ KClO_3(s) \rightarrow KCl(s) + \frac{3}{2} O_2(g) $$

This balances the equation, ensuring that there is one potassium atom, one chlorine atom, and three oxygen atoms on both sides.

How do you balance #KClO_3(s) -&gt; KCl(s) + O_2(g)#?

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