[versyss]

In lab I prepared Mn(acac)3 and I have a pretty general question. Here is how I made the product, in general:

MnCl2*4H2O + NaC2H3O2*3H2O = A

A + C5H8O2 = B

B + KMnO4 = D

D + NaC2H3O2*3H2O = Mn(C5H7O2)3

I know it is way to complicated to figure out what A-D are, so knowing it is an Ox/red reaction I write this overall equation:

Mn+2  + Mn+7  =  Mn+3

So my question is how would I go about finding what the theoretical yield would be, assuming Mn is the Limiting Reagent in all the reactions.  I think I am getting stuck on how I would go about getting the amounts of Mn....

I may just be having a brain fart, but I need some help, Thanks.

[Alberto_Kravina]

I know it is way to complicated to figure out what A-D are, so knowing it is an Ox/red reaction I write this overall equation:

Mn+2  + Mn+7  =  Mn+3

The product of this reaction is Mn⁴⁺ (in neutral - slightly acidic solution).

[Borek]

I know it is way to complicated to figure out what A-D are, so knowing it is an Ox/red reaction I write this overall equation:

Mn+2  + Mn+7  =  Mn+3

The product of this reaction is Mn⁴⁺ (in neutral - slightly acidic solution).

Not exactly - note that versyss wrote just the overall equation that have to show the stoichiometry of the process, not the individual real reaction. As s/he gets Mn(acac)₃ (whatever it is ) final product probably contains Mn(III) - and thus the skeletal equation is OK.

[versyss]

I did end up figuring out my own question!  I needed to balance the charges, quite a simple solution!  After doing that I proceeded to prefrom the typical limiting reagent calculations.  I just tried to overthink it.  Thanks for the help though!

acac is acetylacetone.

[AWK]

If you oxidize Mn(acac)2, the yield depends only on amount of Mn(acac)2 used. You should use an excess of acetylacetone.