Question

A 0,518 g sample of limestone was dissolved and then the calcium was precipitated as calcium oxalate ( CaC2O4). Afer filtrating the precipitate, it was needed 40,0 ml of KMnO4 0,250 N acidified with H2SO4 to titrate it. ¿ What is the %w of CaO in the sample? The equation for the titration is:
MnO-4 + CaC2O4 --> CaSO4 + Mn-2 +CO2 + H2O
The equation is not balanced because I have to use equivalents to get the result.
The result has to be 54,2%
I don't know what to do. I would like to know the steps

Answer 1

1 Ca = 1 CaO = 1 C2O4^2-

2MnO4^- + 5C2O4^2 ==> .....-

mew = milliequivalent weight.
mL x N x mew = grams
Then (grams/mass sample)*100 = ?

mL = 40.0
N = 0.250
mew = 1/2(molar mass CaO)
I get answer of 52.1%. The molar mass CaO may be a little different in your text/notes.

All of the numbers are in the problem except for the mew of CaO.
1 mol Ca = 1 mol CaO = 1 mol C2O4^2-
Titration step is
5 mol C2O4^2- = 1 mol MnO4^-
So 5 mol CaO = 2 mol MnO4^-
which makes
5/2 mol CaO = 1 mol MnO4^0
Now you want to change that to equivalents. The MnO4^- changes by 5 electrons so
(5/2)/5 = 1 eq MnO4^-
and that makes equivalent weight CaO 1/2 molar mass. (i.e., (5/2*5)= 1/2.