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Oxidation of ethanedioate ions by acidified potassium manganate(VII)

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    Hi, I am really stuck on this question about the molar ratios in this question, in fact I am not sure where to start. All I have done is work out the oxidation numbers, anyway here it is:

    In acid solution, potassium manganate(VII) acts as an oxidising agent
    and reacts according to the half-equation:

    5e− + MnO4− + 8H+ → Mn^2+ + 4H2O

    Acidified potassium manganate(VII) will oxidise ethanedioate ions (C2O42−)
    to carbon dioxide.

    What is the mole ratio (MnO4− : C2O4^2−) in which these ions react together?

    Can anyone offer any clues? Thank you!
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    try writing out the oxidation of C2O42− to CO2 with electrons first, then multiply both equations so that the electrons are the same...
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    oh thank you, I finally worked it out to be 2:5 I think..
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    (Original post by escondido)
    oh thank you, I finally worked it out to be 2:5 I think..
    Yup

    2MnO4- + 16H+ + 5C2O42- --> 2Mn2+ + 8H2O + 10CO2
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    Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?
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    (Original post by cpchem)
    Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?
    Yeah I know, I was quite miffed when I got to university since they don't really use the IUPAC names for a lot of things...the trivial and IUPAC versions should both be taught and tested :yep:
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    do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/
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    (Original post by jimmcc)
    do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/
    This is quite an old thread. You'd be better off making a new one asking your question there
Updated: April 13, 2012
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