If x gm is the mass of NaHC2O4 required to neutralize 100 ml of 0.2 (M) NaOH and y gm that is required to reduce 100 ml of 0.02 (M) KMnO4 in acidic medium, then:

Explanation

To solve this problem, we need to consider the stoichiometry of the reactions involved:For neutralizing NaOH, the reaction is: NaHC2O4 + NaOH → Na2C2O4 + H2O. Here, 1 mole of NaHC2O4 neutralizes 1 mole of NaOH.For reducing KMnO4 in acidic medium, the reaction is: 5 NaHC2O4 + 2 KMnO4 + 3 H2SO4 → 5 CO2 + 2 MnSO4 + K2SO4 + 8 H2O. Here, 5 moles of NaHC2O4 reduce 2 moles of KMnO4.From the stoichiometry, it is clear that the mass of NaHC2O4 needed to neutralize NaOH (x) is four times the mass needed to reduce KMnO4 (y), hence x = 4y. Other options do not align with the stoichiometric calculations of these reactions.