Calculate the Percent H2C2O4*2H2O in the unknown sample? Approx. 1mL of H2C2O4*2

Question

Calculate the Percent H2C2O4*2H2O in the unknown sample?
Approx. 1mL of H2C2O4*2H2O liquid is transferes to a 250 mL flask, and add 20 mL DI water. while swirling slowly add 25mL of 3M sulfuric acid, then heat the solution to 70°C-80°C. Tritrate with standardized KMnO4 solution. Calculate the Percent H2C2O4*2H2O in the unknown sample.

M KMnO4 = 0.0076+/-0.0003 g/mol

im not sure if my work is heading in the right direction in calculating % oxalic acid diydrate in sample.

Trial Trial 3 Trial 2 110-03 111.7 11 3559 Vaniins Kea 16.NL 556 aL 6.5mL 444n sars 16. mL 1SmL 2 K M0+5u2H,032M.sa 1so ecor e

Explanation / Answer

Let us do the calcualtion for trial 1 (as you have asked the varification of the same)

We know that amount of sample containing oxalic acid is = 1.796 grams

volume of KMnO4 used = 20.2mL = 0.0202L

Molarity of KMnO4 = 0.0076+/-0.0003 g/mol, i think it should be 0.0076+/-0.0003 moles / L

moles of KMnO4 used = Molarity X volume = 0.0076 X 0.0202 = 0.0001535 moles

the reaction is 2MnO4 + 5H2C2O4 + 6H+2Mn2++10CO2+8H2O

So two moles of KMnO4 reacts with five moles of oxalic acid

therefore 1 mole of KMnO4 will react with 5/2 moles of oxalic acid

0.0001535 moles will react with = 5 X 0.0001535 / 2 moles of oxalic acid = 0.000384 moles of oxalic acid

mass of oxalic acid = Moles X moelcular weight = 0.000384 X 126.07 = 0.0484 grams

% of oxalic acid = 0.0484 X 100 / 1.796 = 2.69%

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