How do you calculate moles to neutralise oxalic acid?

[Daniel2244]

Homework Statement

calculate the number of moles of potassium manganate required to neutralise oxalic acid.
2KMn)₄^- + 5H₂C₂O₄^2-+ 16H⁺ --> 2Mn²⁺ + 10CO₂ + H₂O

Homework Equations

n = c x V
n= m/Ar

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.

 

[Borek]

Just follow the stoichiometry. How many moles of KMnO₄ are needed to oxidize (not neutralize!) 5×10^-4 moles of oxalic acid? In what volume of 0.01085 M solution will you find this amount of permanganate?

There are several problem with your reaction equation, KMnO₄/(COOH)₂ ratio is still OK though.