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Question

0.804 g sample of iron ore was dissolved in acid. Iron was reduced to +2 state and it required 47.2 mL of 0.112 N KMnO4 solution for titration. Calculate the percentage of iron of Fe3O4 in the ore.

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Solution

The titration involves the conversion of ferrou into ferric,
5Fe2++MnO4+8H+5Fe3++Mn2++4H2O
47.2 mL of 0.112 N KMnO447.2 mL of 0.112 N Fe2+ ions
=57.2×0.112×55.51000=0.2934
Mass of iron =0.2934 g
% of iron in the ore =0.29340.804×100=36.49
3Fe(3×55.5)Fe3O4(3×55.5+64)
3×55.5 g of iron form 230.5 g of Fe3O4.
0.2934 g of iron will form =230.5166.5×0.2934=0.406 g

% of Fe3O4 in the ore =0.4060.804×100=50.5.

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