The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine:
3H2O + 3Br2 + I- →6Br- + IO3- + 6H+
The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate:
Ba2+ 1 2IO3+ →Ba(IO3)2
In the analysis of a 1.59-g sample, 0.0538 g of barium iodate was recovered. Express the results of this analysis as percent potassium iodide.
We need at least 10 more requests to produce the solution.
0 / 10 have requested this problem solution
The more requests, the faster the answer.