Sodium oxalate and Potassium Permanganate reaction stoichiometry

Question

An impure sample of sodium oxalate weighing 1.557 g required 27.59 mL of 0.0978 M potassium permanganate for complete reaction according to the following reaction:

5Na2C2O4 + 2KMnO4 + 8H2SO4 = 5Na2SO4 + K2SO4 + 2MnSO4 + 1OCO2 + 8H2O

a) How much pure sodium oxalate was in the sample?

b) What is the percentage of sodium oxalate in the sample?

Answer 1

5Na₂C₂O₄ + 2KMnO₄ + 8H₂SO₄ = 5Na₂SO₄ + K₂SO₄ + 2MnSO₄ + 1OCO₂ + 8H₂O

moles of KMNO₄ needed = 27.59 ml x 1 L/1000 ml x 0.0978 mol/L = 0.002698 moles KMnO₄

moles of Na₂C₂O₄ present = 0.002698 mol KMnO₄ x 5 mol Na₂C₂O₄ / 2 mol KMnO₄ = 0.006746 moles

mass of Na₂C₂O₄ = 0.006746 moles Na₂C₂O₄ x 134 g/mol = 0.9039 g Na₂C₂O₄ = answer (a)

b) percentage Na₂C₂O₄ = mass Na₂C₂O₄ / total mass (x100%) = 0.9039 g/1.557 g (x100%) = 58.06%