In the first titration, the KMnO4 will react with ethanedioate ions. Hence total number of ethanedioate ions from both sodium ethanedioate and ethanedioic acid is found:
n(C2O42-) = 1.33 x 10^-2 mol
In the second titration, the NaOH will react with the ethanedioic acid.
2NaOH + H2C2O4 -> Na2C2O4 + 2H2O
n(H2C2O4) = 5.23 x 10^-3 mol
1.33 x 10^-2 mol = ethanedioate from Na2C2O4 + ethanedioate from H2C2O4
5.23 x 10^-3 mol = ethanedioate from H2C2O4
hence ethanedioate from Na2C2O4 = 1.325 x 10-2 – 5.225 x 10-3 = 8.025 x 10-3 mol
hence number of moles of Na2C2O4 = 8.025 x 10^-3 mol