Concept Explanation: Redox Titration and Stoichiometry

This problem involves determining the quantity of MnCl₂ in a solution through a series of chemical reactions. The key steps are:

1. Step 1: Understand the conversion reaction
   MnCl₂ is converted to KMnO₄ using K₂S₂O₈ and H₂O. The balanced equation for this conversion is: $2\mathrm{MnCl}2+5K2S2O8+8H2O\to 2\mathrm{KMnO}4+6K2\mathrm{SO}4+4H2\mathrm{SO}4+4\mathrm{HCl}$ From this, 2 moles of MnCl₂ produce 2 moles of KMnO₄. So, the mole ratio is 1:1 between MnCl₂ and KMnO₄.
2. Step 2: Analyze the titration with oxalic acid
   The KMnO₄ produced is titrated with oxalic acid (H₂C₂O₄). The reaction in acidic medium (HCl added) is: $2\mathrm{KMnO}4+5H2C2O4+3H2\mathrm{SO}4\to K2\mathrm{SO}4+2\mathrm{MnSO}4+10\mathrm{CO}2+8H2O$ Here, 2 moles of KMnO₄ react with 5 moles of H₂C₂O₄. So, the mole ratio is 2:5 between KMnO₄ and H₂C₂O₄.
3. Step 3: Relate oxalic acid to MnCl₂
   From Step 1, 1 mole MnCl₂ ≡ 1 mole KMnO₄.
   From Step 2, 2 moles KMnO₄ ≡ 5 moles H₂C₂O₄, so 1 mole KMnO₄ ≡ (5/2) moles H₂C₂O₄.
   Therefore, 1 mole MnCl₂ ≡ (5/2) moles H₂C₂O₄.
4. Step 4: Calculate moles of oxalic acid used
   Mass of oxalic acid = 225 mg = 0.225 g.
   Molar mass of H₂C₂O₄ = (2×1 + 2×12 + 4×16) = 90 g/mol.
   Moles of H₂C₂O₄ = mass / molar mass = 0.225 / 90 = 0.0025 mol.
5. Step 5: Find moles of MnCl₂
   From the equivalence: moles of MnCl₂ = moles of H₂C₂O₄ × (2/5) = 0.0025 × (2/5) = 0.001 mol.
6. Step 6: Calculate mass of MnCl₂
   Molar mass of MnCl₂ = 55 + 2×35.5 = 126 g/mol.
   Mass of MnCl₂ = moles × molar mass = 0.001 × 126 = 0.126 g = 126 mg.

Final Answer: The quantity of MnCl₂ present in the initial solution is 126 mg.

Related Topics

• Redox Reactions
• Stoichiometry
• Titration (Specifically, permanganometry)
• Balancing Chemical Equations
• Mole Concept

Important Formulae and Theory

• Mole Concept: Number of moles = mass / molar mass
• Stoichiometry: Use coefficients from balanced equations to find mole ratios.
• Redox Titration: KMnO₄ acts as an oxidizing agent, while oxalic acid is a reducing agent. The endpoint is indicated by the disappearance of the purple color of MnO₄⁻.
• Balancing Equations: Ensure atoms and charges are balanced. In redox reactions, balance the change in oxidation numbers.