$word_{count}$: 230

Reaction: 2KMnO4 + 5H2C2O4 + 3H2SO4 → 2MnSO4 + K2SO4 + 10CO2 + 8H2O. Mole ratio 2:5. n-factors: KMnO4 = 5 (Mn7+ → Mn2+), H2C2O4 = 2 (2C3+ → 2C4+). Self-indicating: purple MnO4- becomes colorless Mn2+ during reaction; endpoint = first persistent pink/purple. Procedure: Oxalic acid in flask + dilute H2SO4 + heat to 60-70 degrees C → add KMnO4 from burette drop by drop. First drops slow (autocatalytic — needs Mn2+ as catalyst). Later drops instant. Endpoint: 30 seconds persistent pink. Why H2SO4? HCl would be oxidized by KMnO4 (Cl- → Cl2). HNO3 is itself an oxidizing agent. H2SO4 is inert to both oxidation and reduction. Why heat? Reaction too slow at RT. But NOT above 70 degrees C (oxalic acid decomposes). Calculations: Use M1V1/n1 = $\frac{M2V2}{n2}$ where n = stoichiometric coefficient. Or N1V1 = N2V2 with N = M x n-factor. Example: 25 mL of 0.1 M oxalic acid needs V mL of 0.02 M KMnO4. Moles oxalic = 2.5 x 10^-3. Moles KMnO4 = $\frac{2}{5}$(2.5 x 10^-3) = 10^-3. V = 10^-3/0.02 x 1000 = 50 mL. JEE favorites: calculate molarity from volume data, identify correct acid, explain self-indicating nature, explain autocatalysis.