Question

5 Fe2+(aq) + MnO4-(aq) + 8 H+(aq) → 5 Fe3+(aq) + Mn2+(aq) + 4 H2O(l) An iron sample weighing 0.379 g is converted into Fe2+(aq) and requires 31.57 mL of MnO4-(aq) according to the equation above. What is the Molarity of the MnO4-(aq) solution?

Answer 1

Mole ratio of Fe²⁺ and MnO₄^- is 5:1 ( from balanced equation).

Moles of Fe²⁺ = Mass of Fe²⁺ / atomic mass of Fe = 0.379 g/(55.845 g/mol) = 0.00679 mol

Moles of MnO₄^- = 0.00679 mol Fe²⁺ x ( 1 mol MnO₄^- / 5 mol Fe²⁺ ) = 0.00136 mol MnO₄^-

Molarity of MnO₄^- = moles of MnO₄^- / volume of MnO₄^- = ( 0.00136 mol/31.57 mL) x ( 1000 mL/ 1 L)

                                       = 0.0431 mol/L

So, the molarity of MnO₄^- solution is 0.0431 mol/L.