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?
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?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd47a22ed-01f7-4417-a334-9ebcd1dbcfe0%2Fba0f01dd-a800-4e76-8220-7d0904d3b541%2Fzdu227_processed.png&w=3840&q=75)
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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