1. The problem statement, all variables and given/known data Bleaching powder, Ca(OCl)Cl, reacts with iodide in acidic medium according to the equation: OCl– + 2 I– + 2 H3O– ---> I2 + Cl– + 3 H2O How many millilitres of 0.0645 M Na2S2O3 are required to titrate the iodine liberated from a 0.6028 g sample of bleaching powder containing 10.50 % (w/w) Cl ? 2. Relevant equations I2 + 2S203 --> 2I- + S4O6 3. The attempt at a solution I have already submitted an answer for this assignment, and this was one of only 2 questions I got wrong, we can do a second submission so I went to figure this one out. I think this is where I went wrong. I Figured out the Cl- moles by: g Cl/0.6028 g = 0.01050 0.063294 g Cl / 35.453 g/mol = 1.78529 mmol I think I may have done that wrong, but from that. From mol Cl- I used 1:1 ratio of original reaction to get mol I2 Mol I2 in second reaction are 1/2 mol S203 mol S2O3 = 3.57058641 mmol Then calculated mL needed. V = 3.57058641 mmol/0.0645 M = 55.4 mL Any suggestions?