Question

A 1.147 gram sample containing an unknown amount of arsenic trichloride and the rest inerts was dissolved into a NaHCO3 and HCl aqueous solution. To this solution was added 1.690 grams of KI and 50.00 mL of a 0.00821 M KIO3 solution. The excess I3– was titrated with 50.00 mL of a 0.02000 M Na2S2O3 solution. What was the mass percent of arsenic trichloride in the original sample?

Answer 1

Note: I don't use I^3-; it's the I2 that does the job.
As^3+ + I2 ==> 2I^- + As^5+
Then mols I2 initially = 0.050L x 0.0821 = 0.004105
mols I2 titrated with S2O3^2- = 0.050 x 0.02 = 0.001 which means 0.0005 mols I2 were used.
I2 + 2S2O3^2- ==> S4O6^2- + 2I^-
mols I2 used for the As/I2 rxn = 0.004105-0.0005 = 0.003605

Then g As = mols As x atomic mass As
%As = (g As/g sample)*100 = ?