Question

A 2.25g sample of impure oxalate requires 30.5 ml of .0210 M KMnO4 for oxidation in acid. what is the percent Na2C2O4?

Answer 1

we know that

moles = conc x volume (ml) / 1000

so

moles of KMnO4 = 0.021 x 30.5 / 1000

moles of KMnO4 = 6.405 x 10-4

now

the reaction is given below

2 KMnO4 + 5 Na2C204 + 8 H2S04 ---> K2S04 + 2 MnS04 + 5 Na2S04 + 10 C02 + 8 H20

we can see that

moles of Na2C204 = 2.5 x moles of KMnO4

so

moles of Na2C2O4 = 2.5 x 6.405 x 10-4

moles of Na2C204 = 1.60125 x 10-3

now

moles = mass x molar mass

so

mass of Na2C204 = 1.60125 x 10-3 x 134

mass of Na2C2O4 = 0.21456 g

now

% mass = mass of Na2C204 x 100 / total mass

% mass = 0.21456 x 100 / 2.25

% mass = 9.536

so

percent Na2C204 in the sample is 9.536%