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# Oxidation of ethanedioate ions by acidified potassium manganate(VII)

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1. 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!
2. try writing out the oxidation of C2O42− to CO2 with electrons first, then multiply both equations so that the electrons are the same...
3. oh thank you, I finally worked it out to be 2:5 I think..
4. (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
5. Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?
6. (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
7. do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/
8. (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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