In titrimetric anlysis of iron ore, the sample is disslved in acid and all the iron were reduced to Fe2+. The Fe2+ are them oxidized to Fe3+ by titrating with a standard solution of an oxidizing agent. A certain iron ore contains 19.7% FeO and 14.3% Fe2)3. What volume of 0.0204M KMnO4 will be required to titrate the iron in a 0.642g sample of the ore? The titration is done in acid solution and the reaction is,Fe2+ + MnO-4 + H+ gt;gt;gt;gt; Fe#+ + Mn2+ + H2O
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Mass FeO 19.7 x 0.642 / 100 0.126 g Mass Fe2O3 14.3 x 0.642 / 100 0.0918 g Moles FeO moles Fe2+ 0.126 g / 71.846 g/mol 0.00175 Moles Fe2O3 0.0918 g / 159.69 g/mol 0.000575 Moles Fe3+ 0.000575 x 2 0.00115 0.00115 moles Fe3+ are reduced to 0.00115 moles Fe2+ Total moles Fe2+ 0.00115 + 0.00175 0.00290 the balanced equation is MnO4- + 5 Fe2+ + 8H+ Mn2+ + 5Fe3+ + 4H2O the ratio between Fe2+ and MnO4- is 5 : 1 Moles MnO4- needed 0.00290 / 5 0.000580 V moles / M 0.000580 / 0.0204 0.0284 L 28.4 mL
