[ve_90]

I need to balance this reaction using the oxidation number change method, which I think is preferable to the half-reaction method, since the reaction is written in molecular form.
K₂CrO₄ + KI + HCl → CrCl₃ + KCl + I₂ + H₂O
I know that chromium is reduced (from +6 to +3) while iodine is oxidized (from -1 to 0).
I also know that I need to balance the iodine atoms by placing a 2 in front of KI on the reactant side.
The balanced reaction, according to the book solution, is as follows:
2K₂CrO₄+ 6KI + 16HCl → 2CrCl₃ + 10KCl + 3I₂ + 8H₂O.
I can't understand how it was solved: can someone help me?

[Hunter2]

Recipe
1. Find the redox pairs
CrO42- / Cr3+  I-/I2
2. Find the condition acidic, neutral or alcaline
If oxygen is involved in acidic reduction add H+ and get H2O, for oxidation vise versa, if neutral or alcaline reduction add water and get OH-.
3. Develop the equation
4. Balance the charges with electrons.
5. Built lowest common multiple of electrons and multiply accordingly
6. Add both equation and eliminate the electrons, H+ , H2O , OH- if necessary.
The redox equation is finished
7. Add the spectator ions, if necessary.

[Borek]

I need to balance this reaction using the oxidation number change method

If Cr gets reduced by 3, and iodine gets oxidized by 1, there must be three KI for every K₂CrO₄ on the left. That should be enough to predict coefficients for the main products and balance everything else. Don't be afraid of using fractions at this stage, you can always multiply the equation by a small integer to get rid of them.