Homework Help: Chemistry College Level

Posted by Tichele on Sunday, April 17, 2011 at 4:52pm.

Ok, so here's my question:
If exactly 14.563 grams of KMnO4 were dissolved to yield 750 mL of solution waht would be the normality of the KMnO4? From a previous question (that this question says to refer to) we have:

MnO4- +8H+ +5e --> Mn2+ +4H2O

I'm not exactly sure what to do. I know that N=#eq/L but that's it...Any help would be lovely.

Chemistry College Level - bobpursley, Sunday, April 17, 2011 at 5:03pm
Well, I see 5 equivalents per molecule of KMnO4

Normality= 5*molarity

calculate molarity and you have it.

You have to be careful on potassium permanganate, in caculating equivalents, because the oxidation change for Mn is usually different in each reaction.
Chemistry College Level - Tichele, Sunday, April 17, 2011 at 7:45pm
Ok, I see.

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