The iodide in a sample that also contained chloride wasconverted to iodate by treatment with an excess of bromine: 3H[2]O + 3Br[2] + I^—> 6Br^- + IO^3- +6H^+ The unused bromine was removed by boiling; an excess of bariumion was then added to precipitate the iodate: Ba^2+ + 2IO[3]^- –>Ba(IO[3])[2] In the analysis of a 1.97 g sample, 0.0612 g of barium iodatewas recovered. Express the results of this analysis aspercent potassium iodide.