[Lourdes]

Hello
I need a little help with this redox equation:
Divalent manganese and hypobromite react in basic solution to form potassium permanganate and bromide.
Mn2 + + BrO- → KMnO4 + Br-
I have balanced with OH and H2O after determining what is being oxidized / reduced.
Ox .: Mn2 + + 8 OH → MnO4 + 4 H2O + 6e-
Red .: BrO- + H2O + 2e- → Br- + 2OH-
I then multiplied the second equation by 3, cancelling out the electrons, and finally writing everything down.
Mn2 + + 8 OH + 3 BrO + 3 H2O → MnO4 + 4 H2O + 3 Br + 6 OH-
Now the question is now: If 27 g of potassium hypobromite react with manganese sulphate, how much g of potassium permanganate is produced?
I find this very funny, because nowhere in my equation is there potassium hypobromite or manganese sulfate, how will I calculate this?
Thanks for your help.

[Hunter2]

Your balance of the equation is not correct. You have to use all charges.
And for the Ions you can write the salts.

[Lourdes]

Sorry, I am a newbie, this is my first time doing redox. Can you help me out here, what you are saying is not so clear to me. 

[Borek]

Balancing redox equation is not only about about balancing atoms (which reflects mass conservation) but also about balancing charge (which is about charge conservation). That means both the total charge nor the number of atoms change during reaction (or half reaction).

Write your half reactions (using our formatting tools above the edit field) so that they are complete, with all involved charges.

Are the charges balanced?

Now the question is now: If 27 g of potassium hypobromite react with manganese sulphate, how much g of potassium permanganate is produced?
I find this very funny, because nowhere in my equation is there potassium hypobromite or manganese sulfate, how will I calculate this?

Where does the BrO^- come from? Where does the Mn²⁺ come from?

[Lourdes]

I am sorry but I have no idea what you guys are talking about. 
I thought all my charges were balanced when I added the electrons. Thanks anyway for your replies!

[Hunter2]

No.

You wrote   Mn2 + + 8 OH → MnO4 + 4 H2O + 6e-

Neither the OH nor the MnO4 have charges.

Correct is:  Mn²⁺  + 8 OH^- → MnO₄^- + 4 H₂O + 5 e^-

[Lourdes]

Oh, that is what you meant, I forgot the charge, sorry 
But other than that, shouldn't they be correct?

[Hunter2]

Yes and by forgotten it, you got a wrong equation. The reduction equation is ok. And BrO^- comes from KBrO and Mn²⁺ comes from MnSO₄.

[Lourdes]

OK, thank you very much. I am now going to try the second part of this redox, hope this works out!

[Lourdes]

OK. Here are my balanced half-reactions:

Ox.: Mn2+ + 8 OH- → MnO4- + 4 H2O + 5e-

Red.: BrO- + H2O + 2e- → Br- + 2 OH-

The first one I multiplied by 2, the other one by 5. Then I cancelled out the electrons and finally:

2 Mn2+ + 6 OH- + 5 BrO- → 2 MnO4- + 3 H2O + 5 Br-

I checked the atoms and the charges, they are the same on both sides.

So now it says, that if I have 27 g of KBrO how much KMnO4 is going to form?
What I understood from your replies was that I take the coefficients that I got from my redox  equation and use them here so that is what I did:

5 KBrO + 2 MnSO4 → 2 KMnO4

The bold ones are the ones that interest me now. So when I calculate how much KMnO4 is going to form by using up 27 g of KBrO I get about 12,64 g.

Is this the correct answer, did I even get it? Because otherwise I have no idea...
I hope someone can help me, thanks

[Hunter2]

The result is correct.

[Lourdes]

Thank you very much, you really helped me out today!