posted by tomi
The following equation describes the oxidation of an acidified solution of sodium oxalate (Na2C2O4) by potassium permanganate(KMnO4)
5C2O4 + 2MnO4 + 16H2SO4 --> 10CO2 + 2Mn + 16HSO4 + 8H20
What volume of 0.127M sodium oxalate in sulphuric acid will react with 3.57g of potassium permanganate?
Convert 3.57 g KMnO4 to moles. mols = grams/molar mass.
Using the coefficients in the balanced equation, convert mols KMnO4 to mols Na2C2O4.
molarity Na2C2O4 = mols/L. YOu know mols from the above calculation, you know M, calculate volume.
Post your work if you get stuck.
M KMnO4 = 158.0339g/mol
3.57g X -----------------
5 mol C2O4-
= 0.02259009 mol KMnO4 X -------------
2 mol KMnO4
=0.056475225 mol C2O4-X ---------------
0.127 mol C2O4-
= 0.445 L
Woops sorry about the spacing. I didn't realize it would come out like that. Hope you can read it.