Redox titration (calculating percent iron in sample)

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Homework Statement

A mass of iron ore weighing 0.2792g was dissolved in dilute acid and all the iron was converted to Fe2+(aq). The iron II solution required 23.30ml of 0.0194M KMnO4 for titration. Calculate the percentage of iron in the ore.

Homework Equations

Wrote out my redox reactions:
MnO4- + 8H+ + 5e- ---> Mn2+ + 4H2O (2)
Fe ---> Fe2+ + 2e- (5)
2MnO4- + 16H+ + 5Fe ---> 2Mn2+ + 8H2O + 5Fe2+

The Attempt at a Solution

0.0194M KMnO4 [0.02330L][5mol Fe/2 mol MnO4][55.85g/1mol Fe] = 0.0631g Fe

0.0631g/0.2792g X 100 = 22.6%Seems like an easy enough question, done several like it before but somethings not right, the correct answer should be 45.3%
Edit: Figured it out, my redox reaction for Fe was wrong, should have been [Fe2+ ---> Fe3+ + e-] which change the moles used in calculations. Still need help on my 2nd question.

 

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Another redox question

Homework Statement

Write required reactions:

Homework Equations

Redox Reaction: IBr + BrO3- ---> IO3- + Br- (in acid)

The Attempt at a Solution

I know the steps to doing these, balance centrals atoms, balance O using H2O, and balancing H using H+. I'm just not sure what to do with the Br in writing the first half reaction.


IBr ---> IO3- (I tried to balance this out by adding Br2 to left side, I know that's not right but I'm just not sure how to balance this otherwise)
BrO3- ---> Br-

After writing these out I did the normal steps but obviously my answer didn't work out. Can someone help with point me in the right direction on balancing Br in first half?