The sulfur in a 0.575 g sample of am organic thiamine was converted to H2S, whic

Question

The sulfur in a 0.575 g sample of am organic thiamine was converted to H2S, which was then absorbed in 15.00 mL of 0.0320 M I2. This amount of I2 was in excess of that required to oxidize the H2S to elemental sulfur; I- was the reaction product. H2S + I2 --> S + 2I- + 2H+. The excess I2 was determined by titration with thiosulfate [S2O3]2- with the reaction products being tetrathionate [S4O6]2- and I-. 26.29 mL of 0.0175 M Na2S203 was required to reduce the excess I3. What was the mass of the sulfur in the original sample?

Explanation / Answer

H2S + I2 ? S + 2 I{-} + 2 H{+}

I2 + 2 S2O3{2-} ? 2 I{-} + S4O6{2-}

I would do it this way, but it is essentially the same as your teacher's. It up to you to decide which, if either, is easier.

(0.02629 L) x (0.0175 mol Na2S2O3/L) x (1 mol I2 / 2 mol Na2S2O3) =
0.0002300375 mol excess I2

(((0.01500 L) x (0.0320 mol I2/L)) - (0.0002300375 mol I2)) x (1 mol H2S / 1 mol I2) x
(1 mol S / 1 mol H2S) x (32.0655 g S/mol) = 0.00802 g S = 8.02 mg S

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