Question

Calculate the amount of sodium oxalate needed to react with 1.5 ml of 0.02M of KMNO4 solution.

Answer 1

Ans. Balanced reaction (under acidic conditions)

8 H2SO4 + 2 KMnO4 + 5 Na2C2O4   →   2 MnSO4 + 10 CO2 + K2SO4 + 5 Na2SO4 + 8 H2O

According to the stoichiometry of balanced reaction, 2 moles KMnO4 completely reacts with 5 moles Na-oxalate.

Now, number of moles of KMnO₄ in given sample =        

                                    (Molarity x Volume in Liter) of KMnO₄ solution

                                    = 0.02 M x 0.0015 mL

                                    = 0.02 mol L^-1 x 0.0015 mL                     [1 M = 1 mol L^-1]

                                    = 0.00003 moles

Thus, number of moles of KMnO₄ = 0.00003 moles

Moles of Na-oxalate required for complete reaction

            2 mol KMnO₄ reacts completely with 5 moles Na-oxalate

         Or, 1 mol              -           -           -     (5/2)

         Or, 0.00003 moles - -- -   -           -    (5/2) x 0.00003 moles

                                                            = 0.000075 moles

Thus, amount of Na-oxalate required = 0.000075 moles

            Mass of Na-oxalate required = moles x molecular mass

                                                = 0.000075 moles x 134.00 g mol^-1

                                                = 0.01005 gram = 10.05 mg