0.804g sample of iron ore was dissolved in acid. Iron was reduced to +2 state and it required 47.2mL of 0.112NKMnO4 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++MnO−4+8H+→5Fe3++Mn2++4H2O 47.2mL of 0.112NKMnO4≡47.2mL of 0.112NFe2+ ions =57.2×0.112×55.51000=0.2934 Mass of iron =0.2934g % of iron in the ore =0.29340.804×100=36.49 3Fe(3×55.5)→Fe3O4(3×55.5+64) 3×55.5g of iron form 230.5g of Fe3O4. 0.2934g of iron will form =230.5166.5×0.2934=0.406g