Question

8H+(aq) + 5Fe2+(aq) + MnO4 -->5Fe3+(aq) + Mn2+(aq) + 4H2O(l)

If 1.25 g of Na2C2O4 was titrated with KMnO4, how many milliliters of 0.2145 M KMnO4 would be needed?

Answer 1

The required reaction here:

MnO₄^- + 8H⁺ + 5e - --> Mn²⁺ + 4H₂O
C₂O₄^2- ----> 2CO₂ + 2e-
multiply first by 2 and second by 5 to get 10 electrons in each, then add,
2 MnO₄^- + 16H⁺ + 5C₂O₄^2- ---> 10CO₂ + 8H₂O + 2Mn²⁺

Thus, 2 mole KMnO₄ reacts with 5 mole NaC₂O₄.

Molar mass of NaC₂O₄ : 133.999 g mol⁻¹

1.25 g NaC₂O₄ = 1.25/133.999 = 0.00933 mol.

0.00933 mol NaC₂O₄ required = 0.00933 x 2/5 = 0.003732 mol KMnO₄.

0.2145 M  KMnO₄ = 0.2145 mol in KMnO₄ in 1000 ml.

0.003732 mol KMnO₄ = 0.003732 x 1000/0.2145 = 17.4 ml.

If 1.25 g of NaC₂O₄ was titrated with KMnO₄, 17.4 ml milliliters of 0.2145 M KMnO4 would be needed.