Question

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.

Answer 1

Answer:

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%

Answer 2

Answer:

Explanation:

Given:

Mass of ore = 1.45 g

Volume of KMnO4 = 21.6 mL

Concentration of KMnO4 = 0.102 M

Equation of the reaction

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

MnO4- is seen to be reduced to Mn2+

Number of moles = concentration × volume

= 0.102 × 21.6 × 10^-3 L

= 0.0022 moles.

By stoichiometry, 5 moles of Fe2+ reactes completely with 1 mole of MnO4-. Therefore, number of moles of Fe2+ = 5 × 0.0022

= 0.0110 moles

Molar mass of Fe2+ = 56.5 g/mol

Mass = number of moles × molar mass

= 56 × 0.011

= 0.616 g of Fe2+

Mass of iron in the iron ore = mass of Fe2+/mass of ore × 100

= 0.616/1.45 × 100

= 42.48 %

= 42.5 %