The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: 3 H20 + 3 Br: +I- →6 Br +103- + 6 H* The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: Ba?- + 2 103 → Ba(10:): In the analysis of a 1.97-g sample, 0.0612 g of barium iodate, Ba(10:):, was recovered. a. How many moles of barium iodate (487.15 g/mol) was recovered? b. From your answer in a, how many moles of I03 is produced? c. From you answer in b, how many moles of I- was converted?

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3. The iodide in a sample that also contained chloride was converted to iodate by treatment
with an excess of bromine:
З Н-0 + 3 Brz + I 6 Br +IO3 + 6 H-
The unused bromine was removed by boiling; an excess of barium ion was then added to
precipitate the iodate:
Ba2+ + 2 103 → Ba(103)2
In the analysis of a 1.97-g sample, 0.0612 g of barium iodate, Ba(I0:)2, was recovered.
a. How many moles of barium iodate (487.15 g/mol) was recovered?
b. From your answer in a, how many moles of IO; is produced?
с.
From you answer in b, how many moles of I was converted?
d. From your answer in d, how many grams of KI (166 g/mol) is present?
e. What is the percentage of KI is in the sample?
Transcribed Image Text:3. The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: З Н-0 + 3 Brz + I 6 Br +IO3 + 6 H- The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: Ba2+ + 2 103 → Ba(103)2 In the analysis of a 1.97-g sample, 0.0612 g of barium iodate, Ba(I0:)2, was recovered. a. How many moles of barium iodate (487.15 g/mol) was recovered? b. From your answer in a, how many moles of IO; is produced? с. From you answer in b, how many moles of I was converted? d. From your answer in d, how many grams of KI (166 g/mol) is present? e. What is the percentage of KI is in the sample?
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