What volume (in mL) of a 0.1 M solution is needed to completely oxidize 100 mg of ferrous oxalate ( — Redox Reactions and Volumetric Analysis Chemistry Question

💡 Solution & Explanation

Moles of $FeC_2O_4 = 100 \times 10^{-3} / 144$ . The n-factor of $FeC_2O_4$ is 3 (Fe: +2 to +3; C: +3 to +4). Equivalents of $FeC_2O_4 = 3 \times (0.1 / 144)$ . For $KMnO_4$ in acidic medium, n-factor = 5. Equivalents of $KMnO_4 = 0.1 \times V(in L) \times 5 = 0.5V$ . Equating: $0.5V = 0.3/144 \Rightarrow V = 0.6/144 L = 600/144$ mL = 4.16 mL $\approx$ 4.1 mL.