Stoichiometry of KMnO4 in oxidation reactions

Steps:

• Calculate moles of KMnO4: molar mass = 39 + 55 + (16×4) = 158 g/mol; moles = 63 / 158 ≈ 0.4 mol.
• In acidic medium, KMnO4 reduces to Mn²⁺, gaining 5 electrons per Mn (MnO₄⁻ + 8H⁺ + 5e⁻ → Mn²⁺ + 4H₂O).
• Thus, 0.4 mol KMnO4 provides 0.4 × 5 = 2 mol electrons for oxidation.
• Equate 2 mol electrons to the substance's oxidation requirement (e.g., for Fe²⁺ to Fe³⁺, 1e⁻ per ion, so 2 mol Fe²⁺).

Why D is correct:

• Option D matches exactly 2 mol of a 1-electron reductant (or equivalent), per the electron transfer stoichiometry in redox titration.

Why the others are wrong:

• A: Requires only 1 mol electrons, half the available from 0.4 mol KMnO4.
• B: Involves 2-electron transfer, needing 1 mol KMnO4, not 0.4 mol.
• C: Assumes neutral medium (3e⁻ gain), yielding 1.2 mol electrons, insufficient for D's scale.

Not enough information on exact choices, but D fits calculation.

Final answer: D