Answer to Question #286146 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 (1): Fe₃O₄ → 3Fe³⁺→ 3Fe²⁺

Balanced chemical equation (2): 5Fe²⁺ + MnO₄⁻ + 8H⁺ → 5Fe³⁺ + Mn²⁺ + 4H₂O

Calculate the moles of MnO₄⁻:

KMnO₄ ⇌ K⁺ + MnO₄⁻

Moles of KMnO₄ = Moles of MnO₄⁻

Moles of KMnO₄ = Molarity of KMnO₄ × Volume of KMnO₄ solution

Moles of MnO₄⁻ = Moles of KMnO₄ = (0.025 M × 0.0475 L) = 0.0011875 mol MnO₄⁻

According to the equation (2):

1 mol of MnO₄⁻ reacts with 5 moles of Fe²⁺

Thus, 0.0011875 mol of MnO₄⁻ reacts with:

Moles of Fe²⁺ = (0.0011875 mol MnO₄⁻) × (5 mol Fe²⁺ / 1 mol MnO₄⁻) = 0.0059375 mol Fe²⁺

According to the equation (1):

1 mol of Fe₃O₄ produces 3 mol of Fe²⁺

Thus, X mol of Fe₃O₄ produces 0.0059375 mol of Fe²⁺

Moles of Fe₃O₄ = (0.0059375 mol Fe²⁺) × (1 mol Fe₃O₄ / 3 mol Fe²⁺) = 0.001979 mol Fe₃O₄

Calculate the mass of Fe₃O₄:

The molar mass of Fe₃O₄ is 231.54 g/mol

Hence,

Mass of Fe₃O₄ = (0.001979 mol Fe₃O₄) × (231.54 g Fe₃O₄ / 1 mol Fe₃O₄) = 0.458 g Fe₃O₄

Calculate the %Fe₃O₄:

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

%Fe₃O₄ = (0.458 g / 2.00 g) × 100% = 22.9%

%Fe₃O₄ = 22.9%

Answer: %Fe₃O₄ = 22.9%