The iron content of iron ore can be determined by titration with a standard KMnO4 solution. The iron ore is dissolved in HC1, and all the iron is reduced to Fe²+ ions. This solution is then titrated with KMnO4 solution, producing Fe³+ and Mn²+ ions in acidic solution. If it required 34.08 mL of 0.0510 M KMnO4 to titrate a solution made from 1.957 g of iron ore, what is the mass percent of iron in the iron ore? Mass percent = %
Molarity of KMnO4 = 0.0510 M
Volume of KMnO4 = 34.08 ml = 0.03408 L
We have , No of moles of KMnO 4 = Molarity x Volume of solution in L
No of moles of KMnO 4 = 0.0510 mol / L x 0.03408 L
No of moles of KMnO4 = 17.1 x 10 -04 mol
Consider a reaction : MnO 4- + 5 Fe 2+ + 8 H + 
Mn 2+ + 5 Fe 3+ + 4 H2O
Stoichiometric ratio = No of moles of MnO 4- / No of moles of Fe 2+ = 1/ 5
We can use relation , No of moles of MnO 4- = No of moles of Fe 2+ x Stoichiometric ratio
Therefore, No of moles of Fe 2+ = No of moles of MnO 4- / Stoichiometric ratio
No of moles of Fe 2+ = 17.1 x 10 -04 mol / ( 1/5)
No of moles of Fe 2+ = 8.55 x 10 -03 mol
Mass of Fe in sample = 8.55 x 10 -03 mol x ( 55.84 g / 1 mol ) = 0.477 g
% of Fe in sample = Mass of Fe / Mass of sample x 100
% of Fe in sample = 0.477 g / 1.957 g ) x 100
% of Fe in sample = 24.4 %
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