1. The problem statement, all variables and given/known data 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.