Balancing an equation

[LoLPoPs]

When balancing an equation, I usually write half reactions. However, for the reaction below, the elements are mixed in the product...and I'm confused how you would balance it.

KMnO4 + NH3 -> KNO3 + MnO2 + KOH + H2O


Can someone help with this question? Thanks!!!

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[slm266]

When balancing an equation, I usually write half reactions. However, for the reaction below, the elements are mixed in the product...and I'm confused how you would balance it.

KMnO4 + NH3 -> KNO3 + MnO2 + KOH + H2O


Can someone help with this question? Thanks!!!

I've never seen problems like that, especially not on the dat.

 

[flyingforever]

When balancing an equation, I usually write half reactions. However, for the reaction below, the elements are mixed in the product...and I'm confused how you would balance it.

KMnO4 + NH3 -> KNO3 + MnO2 + KOH + H2O


Can someone help with this question? Thanks!!!

Whats Mn and O charges? Doesn't seems right though. I think it should be MnO (having +2 for Mn and -2 for O).

 

[UndergradGuy7]

Still can find the 2 half reactions...

Just find the main elements that are oxidized/reduced. K can never be oxidized its +1. Here if you check N changes oxidation state. So does the MnO4. Even if you don't know what MnO2 is. You don't need to know to balance it I think. (I think MnO2, Mn is +4.)

So...
NH3 -> NO3^-
MnO4^- -> MnO2
Ignore everything else and the K. K is a spectator ion.

Then do the usual balancing redox. Balance it for a "basic solution" since we have KOH on the right. Add H2O for O, etc...

Once you get to the end you will get
8MnO4^- + 3 NH3 ---> 8MnO2 + 3NO3^- + 2H2O + 5OH^-
and you will wonder, where is the K+?
Just add it in for the MnO4^-, OH^-, and NO3^- (pretty much to every ion). Its a spectator ion so it can be left out, but the answer probably wants it.

8 KMnO4 + 3 NH3 ---> 8MnO2 + 3KNO3 + 2H2O + 5KOH