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:
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 %
