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The iron in a 0.5973g iron ore sample is dissolved and converted to Fe2+. This solution...

The iron in a 0.5973g iron ore sample is dissolved and converted to Fe2+. This solution is titrated with a 0.02151F solution of KMnO4 by the reaction:

5Fe2+(aq) + MnO4-(aq) + 8H+(aq)----- 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

The titration requires 33.23mL of the standard. Determine the percentage by weight of Fe2O3 (MM=159.69g/mol) in the sample.

The correct answer is 47.77% please show all your process. Thanks!

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Answer #1

5Fe2+(aq) + MnO4-(aq) + 8H+(aq) -------> 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

Stoichiometrically, 1 mole of MnO4- reacting with 5mole of Fe2+

Concentration of KMnO4 = 0.02151F

Concentration of MnO4- = 0.02151F

No of mole of MnO4- consumed = (0.02151mol/1000ml)×33.23ml = 0.0007148

0.0007148mole of MnO4 represents 5× 0.0007148= 0.003574 mole of Fe2+

Molar mass of Fe = 55.845g/mol

Mass of Fe2+ = 0.003574mol × 55.845g/mol = 0.19958g

Molar mass o Fe2O3 = 159.69g/mol

Fe2+ to Fe2O3 conversion factor = 159.69g/111.69g=1.4298

Mass of Fe2O3 = 1.4298 × 0.19958g = 0.28535g

% of Fe2O3 = (0.28535g/0.5973g)×100 = 47.77℅

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