calculate the number of moles of potassium manganate required to neutralise oxalic acid.
2KMn)4- + 5H2C2O42-+ 16H+ --> 2Mn2+ + 10CO2 + H2O
n = c x V
The Attempt at a Solution
Moles of oxalic acid = 5x10-4
Volumeol of KMnO4 = 0.01085dm-3
Concentration of oxalic acid = 0.05 therefore concentration of manganate is the same (might be wrong).
moles manganate= c x V = 0.05 X 5x10-4= 5.425x10-4.
However, I think this is wrong and that you're meant to use the mole ratio difference to get the moles, but I'm not sure how.