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Chemistry College Level

posted by on .

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 - ,

    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 - ,

    Ok, I see.

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