A 1.45 g sample of an iron ore is dissolved in acid and the iron obtained is Fe2+(aq). To titrate the solution, 21.6 mL of 0.102 M KMnO4(aq) is required. What is the percent of iron in the ore? Hint: calculate moles of iron then mass of iron then % iron.

42.4%

Explanation:

5Fe2+(aq) + MnO4^- (aq) + 8H^+(aq)> 5Fe^3+(aq) + 4H2O(l) + Mn2+(aq)

Concentration of oxidizing agent =0.102M

Volume of oxidizing agent= 21.6ml

Amount of MnO4^- reacted= CV= 21.6ml/1000 × 0.102= 2.2 × 10^-3moles

From the balanced redox reaction equation

5 moles of Fe^2+ required 1 mole of MnO4^-

x moles of Fe^2+ will require 2.2 ×10^-3 moles of MnO4^-

x= 5×2.2×10^-3/1 = 11 ×10^-3moles of Fe^2+

Therefore, mass of Fe^2+ reacted = 11×10^-3 × 56= 616×10^-3g of Iron

% 0f iron present = 0.616/1.45 × 100= 42.4%

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