[escondido]

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!

[EierVonSatan]

try writing out the oxidation of C₂O₄²⁻ to CO₂ with electrons first, then multiply both equations so that the electrons are the same...

[escondido]

oh thank you, I finally worked it out to be 2:5 I think..

[EierVonSatan]

(Original post by escondido)
oh thank you, I finally worked it out to be 2:5 I think..

Yup

2MnO₄^- + 16H⁺ + 5C₂O₄^2- --> 2Mn²⁺ + 8H₂O + 10CO₂

[cpchem]

Urgh. Nasty IUPAC-ness. What is wrong with the name oxalate?

[EierVonSatan]

(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

[jimmcc]

do you know the order of reaction with regards to all the reactants? I'm pretty stuck and in desperate need of help :/

[Ari Ben Canaan]

(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