I am pretty good at balancing oxidation reduction problems, but I can't do this one:
HIO3 + FeI2 + HCL -> FeCl3 + ICl + HOH
5HIO3 + 4FeI2 + 25HCl -> 4FeCl3 + 13ICl + 15HOH
I separated the three redox equations, added the two oxidation reactions together, then balanced that net equation with the reduction (all ionic of course), then added the spectator ions.
Check it to make sure it is correct.
To check the balanced equation you provided for the reaction:
HIO3 + FeI2 + HCl -> FeCl3 + ICl + H2O
First, let's balance the oxidation-reduction half-reactions separately:
Oxidation half-reaction:
HIO3 -> ICl
The iodine is going from an oxidation state of +5 in HIO3 to -1 in ICl, so we need to balance the number of iodine atoms. Since there is only 1 iodine on both sides, this half-reaction is already balanced.
Reduction half-reaction:
FeI2 -> FeCl3
The iron is going from an oxidation state of +2 in FeI2 to +3 in FeCl3, so we need to balance the number of iron atoms. Since there is 1 Fe on both sides, this half-reaction is already balanced.
Now let's combine the two half-reactions:
HIO3 + 2FeI2 + 6HCl -> 2FeCl3 + ICl + 3H2O
Next, let's balance the hydrogen atoms by adding H+ ions:
HIO3 + 2FeI2 + 6HCl -> 2FeCl3 + ICl + 3H2O + 6H+
Finally, let's balance the charge by adding electrons (e-) to the side with the more positive charge:
HIO3 + 2FeI2 + 6HCl -> 2FeCl3 + ICl + 3H2O + 6H+ + 6e-
Now both sides of the equation are balanced in terms of atoms and charge. However, it appears that the coefficients you provided in your initial attempt were incorrect.
Keep in mind that this is just one possible balanced equation for the given reaction. There could be other valid balanced equations, but this is the result of following the steps for balancing oxidation-reduction reactions.