Chemistry Question - gas equation help?

Question

20.00cm^3 of KI solution of unknown concentration is titrated with 0.0250M KMnO4 solution in the presence of excess acid. 14.80cm^3 of the KMnO4 solution is required for complete reaction. Calculate the concentration of the KI solution.
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My workings out so far:
n = cV

first part:
n(KMnO4) = 0.0250 x 14.80 = 0.37
c(KI) = 0.37/20 = 0.0185

The answer has to be 0.0925M
If you divide 0.37 by 4 then the answer comes out, but I'm not sure how to do it. I'd really appreciate some help!!!

Answer 1

First you must balance the chem. equation:

10KI + 2KMnO4 + 8H2SO4 --> 5I2 + 6K2SO4 + 2MnSO4 + 8H2O

The quantity of KMnO4 equals:

n = C*V = 0.025*14.8*10^(-3) = 3.7x10^(-4) mol.

2 mol of KMnO4 react with 10 mol of KI
3.7x10^(-4) mol of KMnO4 react with 5*3.7x10^(-4) = 1.85x10^(-3) mol KI

Now find the molarity of KI solution:

C = n/V, C = 1.85x10^(-3)/20x10^(-3) = 0.0925 M