Please explain how to do this question!

Question

roberts.spurs19 · Answer

13.2 cm^3 of potassium iodate(v) solution was added to an excess of acidified potassium iodide.This solution was then titrated against a 0.200 mol dm^-3 solution of sodium thiosulfate. 41.1 cm^3 of thiosulfate solution was required to react fully with the solution. The equation for the reaction is as follows:
I2 + 2 Na2S2O3 = Na2S4O + 2 NaI

I have already calculated the moles of sodium thiosulfate used and the moles of iodine formed by adding the potassium iodate(v) to the potassium iodide.
Thank you

somy · Answer

What do you want to find here?

roberts.spurs19 · Answer

Sorry! how can I find the number of moles of potassium iodate present please?

somy · Answer

you found mole of Iodide?

vincent-lyon.fr · Answer

You need to write down the equation of the reaction between iodate and iodide ions, then work out the amounts in that first reaction.

somy · Answer

ok, apparently this question is from some book or past paper 
can u give me the link? or take a pic of the whole thing

roberts.spurs19 · Answer

Sorry didn't have internet access last couple of days. 
Thank you for your help I understand now.