A sample of 1.55 g of iron ore is dissolved in an acid solution in which the iron is converted into Fe2+.

Question

The solution formed is then titrated with KMnO4 which oxidises Fe2+ to Fe3+ while the MnO4 - ions are reduced to Mn2+ ions. 92.95 mL of 0.020 M KMnO4 is required for the titration to reach the equivalence point.

a) Write the balanced equation for the titration.

b) Calculate the percentage of iron in the sample.

Answer 1

a) 5Fe²⁺ + MnO₄^- + 8H⁺ ==> 5Fe³⁺ + Mn²⁺ + 4H₂O ... balanced equation

b) mols MnO₄^- used = 92.95 ml x 1 L/1000 ml x 0.020 mol/L = 0.001859 mols MnO₄^- used

mols Fe²⁺ present = 0.001850 mols MnO₄^- x 5 mols Fe²⁺ / 1 mol MnO₄^- = 0.009295 mols Fe²⁺

mass Fe present in original sample = 0.009295 mols Fe x 55.85 g Fe/ mol = 0.5191 g Fe

% Fe in the sample = 0.5191 g / 1.55 g (x100%) = 33.49% = 33% (2 sig. figs.)