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Homework Help: Redox titration (calculating percent iron in sample)

  1. Apr 18, 2010 #1
    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.

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

    3. 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.
    Last edited: Apr 18, 2010
  2. jcsd
  3. Apr 18, 2010 #2
    Another redox question

    Write required reactions:

    2. Relevant equations
    Redox Reaction: IBr + BrO3- ---> IO3- + Br- (in acid)

    3. 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 thats 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?