The iron content of iron ore can be determined by titration with KMnO₄.

Question

8H⁺(aq) + MnO₄⁻(aq) + 5Fe²⁺(aq) → 5Fe³⁺(aq) + Mn²⁺(aq) + 4H₂O(l) A sample of iron ore with a mass of 1.535 g was dissolved with HCl and then treated with a reducing agent to convert any Fe³⁺ in the ore to Fe²⁺. Titration with 0.0455 M KMnO₄ solution required 59.15 mL of the solution to reach the endpoint where the purple MnO₄⁻ ion was no longer being consumed. The molar mass of Fe is 55.85 g/mol. Determine the mass percent iron in the ore

Answer 1

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

moles KMnO₄ used = 59.15 ml x 1 L / 1000 ml x 0.0455 mol / L = 2.691x10^-3 moles

moles Fe²⁺ = 2.691x10^-3 moles MnO4^- x 5 mol Fe2+ / mol MnO4^- = 0.01346 mols

Grams Fe = 0.01346 mols x 55.85 g / mol = 0.7516 g

Mass % Fe = 0.7516 g / 1.535 g (x100%) = 48.96%