Question

An ore is to be analyzed for its iron content by an oxidation-reduction titration with permanganate ion. A 4.230 g sample of the ore is dissolved in hydrochloric acid and passed over a reducing agent so that all the iron is in the form Fe2+. The Fe2+ (aq) is completely oxidized by 31.60 mL of a 0.05120 M solution of KMnO4. The unbalanced equation for the reaction is:

KMnO4(aq) + HCl(aq) + FeCl(aq) ---> MnCl(aq) + FeCl3(aq) + H2O(1)

the balanced equation is:
8H+ + MnO4 + 5Fe2+ â Mn2+ + 5Fe3+ + 4H2O

Required:
Calculate the amount of iron in the sample and its mass percentage in the ore.

Answer 1

See explanation

Step-by-step explanation:

The balanced redox reaction equation is;

8H+ + MnO4^- + 5Fe2+ ---------> Mn2+ + 5Fe3+ + 4H2O

Amount of KMnO4 reacted = 31.60/1000 * 0.05120 = 1.62 * 10^-3 moles

From the reaction equation;

1 mole of MnO4^- reacted with 5 moles of Fe2+

1.62 * 10^-3 moles will react with 1.62 * 10^-3 moles * 5/1 = 8.1 * 10^-3 moles

Mass of Fe2+ reacted = 8.1 * 10^-3 moles * 56 g/mol

Mass of Fe2+ reacted = 0.45 g

Amount of iron in the sample = 0.45 g

Percentage of iron in the sample;

0.45 g/4.230 g * 100 = 10.6 %