Question

if 28.70 ml of 0.0200M KMNO4 was required to titrate a 0.250g sample of K2C2O4 what is the % of C2O4^2- in the oxalate?

Answer 1

The number of moles of KMnO₄ = molarity * volume (lit) = 0.02 * 0.0287 = 0.000574moles

2 MnO4- (aq ) + 5 C2O4-2 (aq ) + 16 H+ (aq ) ----> 2 Mn2+ (aq) + 10 CO2 (g ) + 8 H2O (l)

from the equation

2 moles of KMnO₄ reacts with 5 moles of oxalate

then 0.000574 moles of KMnO₄ reacts with (5 * 0.000574) / 2 = 0.001435 moles of oxalate is titrated.

weight of oxalate titrated = number of moles * formula mass of oxalate = 0.001435 * 88.019 = 0.126 gr of oxalate

the % of oxalate = (0.126 / 0.250) * 100 = 50.523% of oxalate is present