Question

What is the percent by mass of sodium oxalate in a 0.614 g sample that took 20.96 mL of 0.03513 M potassium permanganate to titrate to a light pink endpoint?

Answer 1

so Reaction between sodium oxalate and potassium permanganate is given by fionic equation) 2 mnoun + 560.4? + 8H7 - 2m22t+ 100, +8H20 so the molar ratio for the reaction of KMnOu and G20v2 is 2:5 moles of KMnOu used = Molarity & volume in = 0.03513 MX 20.96mb x 14 1000 m2 - 7.363 x10-4 mois ". Moles of Nal%u in sample = 98 x 70 363x104 mol = 1.841710-3 mol. Molar mass of sodium oxalate = 134 g/mol. & Marsof Nag Caou in sample = 20841x10 s mo) x 1349 /omel = 0.2467 9. mass percent of Na₂C2O4 = 0.2467 g +100 0.6149* = 40.18%