Answer to Question #286147 in Chemistry for khryst

Question

   A 2.00 g sample of an iron ore is dissolved in acid and then reduced to Fe²⁺. It was then titrated against 0.025 M KMnO₄ solution and used 47.5 mL to reach the end point. Calculate the %Fe₃O₄ (231.54 g/mol) of the ore sample.

Answer 1

Solution:

Balanced chemical equation: Fe₃O₄ → 3Fe²⁺

1 Fe₃O₄ ≡ 3 Fe²⁺

Balanced chemical equation: 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

5 Fe²⁺ ≡ 1 MnO₄⁻

Therefore,

5 Fe₃O₄ ≡ 15 Fe²⁺ ≡ 3 MnO₄⁻

and

stoichiometric ratio = 5 mol Fe₃O₄ / 3 mol MnO₄⁻

Thus,

(0.0475 L KMnO₄) × (0.025 mol KMnO₄ / 1 L KMnO₄) × (5 mol Fe₃O₄ / 3 mol KMnO₄) = 0.002 mol Fe₃O₄

The molar mass of Fe₃O₄ is 231.54 g/mol

Therefore,

(0.002 mol Fe₃O₄) × (231.54 g Fe₃O₄ / 1 mol Fe₃O₄) = 0.463g Fe₃O₄

%Fe₃O₄ = (Mass of Fe₃O₄ / Mass of sample) × 100%

%Fe₃O₄ = (0.463 g / 2.00 g) × 100% = 23.15%

%Fe₃O₄ = 23.15%

Answer: %Fe₃O₄ = 23.15%