Answer to Question #202765 in Chemistry for Ra Rayo

Question

The amount of iron in a meteorite is determined by a redox titration using KMnO4 as the titrant. A 0.4185-g sample is dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+ using a Walden reductor. Titrating with 0.02500 M KMnO4 requires 41.27 mL to reach the endpoint. Determine the %w/w Fe2O3 in the sample of meteorite.

Answer 1

Solution:

Balanced chemical equation for the titration reaction:

5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

By stoichiometry, n(Fe²⁺)/5 = n(MnO₄⁻)

n(MnO₄⁻) = C(MnO₄⁻) × V(MnO₄⁻) = 0.02500 M × 0.04127 L = 0.00103175 mol

Therefore,

n(Fe²⁺) = 5 × n(MnO₄⁻) = 5 × 0.00103175 mol = 0.00515875 mol

Fe₂O₃ → 2Fe³⁺ → 2Fe²⁺

According to the scheme above: n(Fe₂O₃) = n(Fe²⁺)/2

n(Fe₂O₃) = n(Fe²⁺) / 2 = 0.00515875 mol / 2 = 0.002579 mol

The molar mass of Fe₂O₃ is 159.69 g/mol.

Therefore,

(0.002579 mol Fe₂O₃) × (159.69 g Fe₂O₃ / 1 mol Fe₂O₃) = 0.4119 g Fe₂O₃

Mass of Fe₂O₃ = 0.4119 g

%w/wFe₂O₃ = (mass of Fe₂O₃ / mass of sample) × 100%

%w/wFe₂O₃ = (0.4119 g / 0.4185 g) × 100% = 98.42%

%w/wFe₂O₃ = 98.42%

Answer: %w/wFe₂O₃ = 98.42%