[rhyno-of-the-port]

Hi all, i have done a synthesis and part analysis of said compound, but am having problems working out the % ethanedioate in my sample. My yield was 5.6244g and it was titrated against 0.01979M KMnO_4 (ave. titre- 20.25cm) anyone help me in the right direction?

[AWK]

May be a synonym - oxalate helps you.

Write down a balanced readox reaction

[rhyno-of-the-port]

So, without a redox reaction, it cannot simply be calculated by finding the FW of oxalic acid, and dividing this by the FW of the potassium trisethanedioatealuminate and multiplying it by 100%?


Another thing, i need to find out how many water molecules are attached to this, very confusing

[Arkcon]

Another thing, i need to find out how many water molecules are attached to this, very confusing

If you have a hunk of crystals, and can heat without decomposing the compound, you can drive off the water, and weigh the loss.  Unless you're hungry for more titration, and can do a Karl Fisher titration.  But I've never been that hungry for a titration.

[AWK]

Hint (unbalanced reaction)
K3AlC6O12 + KMnO4 + H2SO4 = K2SO4 + Al2(SO4)3 + MnSO4 + CO2 + H2O

[rhyno-of-the-port]

If you have a hunk of crystals, and can heat without decomposing the compound, you can drive off the water, and weigh the loss.  Unless you're hungry for more titration, and can do a Karl Fisher titration.  But I've never been that hungry for a titration.
[/quote]
The question just asks what the x is before the .H2O, we weren't told to do any further titrations, i'm clueless

Hint (unbalanced reaction)
K3AlC6O12 + KMnO4 + H2SO4 = K2SO4 + Al2(SO4)3 + MnSO4 + CO2 + H2O

i know i'm being stupid but it's sunday morning lol so once i balance the equation what does it tell me and how do i use it?