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In order to determine the amount of MnCl₂ in an aqueous solution, it was fully converted into KMnO₄ using the reaction: MnCl₂ + K₂S₂O₈ + H₂O → KMnO₄ + H₂SO₄ + HCl (unbalanced equation). A few drops of concentrated HCl were added, and the solution was gently heated. Then, 225 mg of oxalic acid was gradually introduced until the purple color of the permanganate ion vanished. Calculate the mass of MnCl₂ (in mg) originally present in the solution. (Atomic masses in g mol⁻¹: Mn = 55, Cl = 35.5)
- 126
- 225
- 90
- 180
Correct answer: 126
Solution
The reaction converts MnCl₂ to KMnO₄, and the amount of oxalic acid used indicates the moles of KMnO₄ formed. Using stoichiometry, the mass of MnCl₂ originally present is calculated to be 126 mg based on the balanced reaction and molar masses.
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