[Borek]

You don't have to use 6 & 3, to cancel electrons it is enough to multiply first reaction by 2.

However, let's assume I have not told you multiplying by 2 will do. Write reaction as you think you should - that is, multpiplying by 6 & 3.

[Evaldas]

KClO₃ + CrCl₃ + KOH  6K₂CrO₄ + 3KCl + H₂O Doesn't make any sense

[Borek]

But you have not followed the rule. You should multiply equations on both sides. That means 6 CrCl₃ on the left and 6 K₂CrO₄ on the right. Do the same to chlorate and chlorides.

Note: you are trying to balance full equation, that means you will need to spend substantial amount of time trying to balance spectators.

[Evaldas]

3KClO₃ + 6CrCl₃ + KOH  6K₂CrO₄ + 3KCl + H₂O

[Borek]

Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl₃ are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO₃^- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).

[Evaldas]

Now, you need to balance everything else - that is, oxygen, hydrogen and potassium. Do it with KOH and water, don't change coefficients in compounds that are getting reduced or oxidized, as they are already in correct ratio.

Unfortunately after that - as you are trying to balance full reaction equation - you also have to balance chlorine. However, chlorides from CrCl₃ are just spectators, so they don't take part in the redox process. You may treat them as separate from chlorides that are being produced in chlorate reduction (imagine you have started not with chromium chloride, but with chromium nitrate - then you will have to balance NO₃^- as spectator and it will not confuse you by being identical with chlorides produced in the reduction process).

I can't manage to do this... I'm lost.

[Borek]

I suppose chlorides confuse you. So let's get rid of those that are not involved in the redox part of the reaction.

Try now:

3KClO₃ + 6CrBr₃ + KOH  =  6K₂CrO₄ + 3KCl + KBr + H₂O

Just remember - don't touch 3KClO₃, 6CrBr₃, 6K₂CrO₄ and 3KCl - these are already balanced.

[Evaldas]

Ok:
3KClO₃ + 6CrBr₃ + 30KOH = 6K₂CrO₄ + 3KCl + 18KBr + 15H₂O ?

[Borek]

1. Replace now all Br by Cl - just a moment ago we did the reverse to not confuse you, time to get back to the original equation.

2. There is something wrong with this equation - coefficients can be smaller. Do you see it?

[Evaldas]

1. 3KClO₃ + 6CrCl₃ + 30KOH = 6K₂CrO₄ + 3KCl + 18KCl + 15H₂O
2. Yes, I see that it can be divided by 3:
3KClO₃ + 6CrCl₃ + 30KOH = 6K₂CrO₄ + 3KCl + 18KCl + 15H₂O /:3 =>
KClO₃ + 2CrCl₃ + 10KOH = 2K₂CrO₄ + KCl + 6KCl + 5H₂O

[Evaldas]

But see now, I don't get one thing: we have to write that number of electrons as a coefficient on both sides, but how do I act in this situation:
P⁰(s) + O⁰₂(g)  P⁺⁵₂O^-2₅(g)
P⁰  P⁺⁵ + 5e       | 5 | 4 |
O⁰₂ + 2x2e  2O^-2 | 4 | 5 |
So I write 4 next to solid phosphorus and a 5 next to oxygen, but obviously I won't write a 4 neither a 5 next to phosphorus(V) oxide, right?
4P⁰(s) + 5O⁰₂(g)  2P⁺⁵₂O^-2₅(g)
How can this be explained, when does this rule apply?

[Borek]

There are two P atoms in a product molecule, so it will be easier to write phosphorus part as

2P⁰ -> 2P⁺⁵ + 10e^-

(you did the same for oxygen, just the other way around).

[Evaldas]

But wouldn't it then be:
2P⁰ -> 2P⁺⁵ + 10e^-  | 10 | 5 | 2 |
O⁰₂ + 2x2e- -> 2O^-2 |  4 | 2 | 5 |
So accordingly I'd have to write:
2P⁰(s) + 5O⁰₂(g)  ?P⁺⁵₂O^-2₅(g)
Or not? Or for some reason I'm supposed to know that the five will go to oxygen, the two to phosphorus(V) oxide, and then accordingly I'd have to count that I have to write 4 next to phosphorus?

[Borek]

2P⁺⁵ is equivalent to P⁺⁵₂ which you have in the oxide molecule.

[Evaldas]

Ok. Let's go through another reaction 
Cr₂⁺³(SO₄)₃ + Br₂⁰ + NaOH  Na₂Cr⁺⁶O₄ + 3NaBr^- + Na₂SO₄ + H₂O
Br₂⁰ + 2x1e^-  2Br^-     | 2 | 1 | 3
Cr₂⁺³ - 2x3e^-  2Cr⁺⁶ | 6 | 3 | 1
So you said that I should follow the rules and write on both sides the numbers, so I'd have to write on both sides 3s, and I don't need to write 1s:
Cr₂(SO₄)₃ + 3Br₂ + NaOH  Na₂CrO₄ + 3NaBr + Na₂SO₄ + H₂O
But now I don't have a choice but to write a six next to NaBr instead of 3. IS IT OK TO DO SO?