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: A 3.455-g sample of a mixture was analyzed for barium

Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro ISBN: 9780321809247 1

Solution for problem 107 Chapter 4

Chemistry: A Molecular Approach | 3rd Edition

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Chemistry: A Molecular Approach | 3rd Edition | ISBN: 9780321809247 | Authors: Nivaldo J. Tro

Chemistry: A Molecular Approach | 3rd Edition

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Problem 107

107: A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aqueous solution of the sample. The resultant reaction produced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample?

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to calculate the mass percentage of barium in the given sample. Step1: The reaction between barium ion and sulfuric acid takes place in the following manner: Ba (aq) + H SO (aq) --------> BaSO (aq) + 2H (aq) + 2 4 4 From the above equation it is clear that 1 mol of barium sulfate is produced from 1 mol of barium ion. Step2: Given, mass of barium sulfate = 0.2815 g Molar mass of barium sulfate = 233.4 g/mol Therefore, number of moles of barium sulfate = given mass /...

Step 2 of 3

Chapter 4, Problem 107 is Solved
Step 3 of 3

Textbook: Chemistry: A Molecular Approach
Edition: 3
Author: Nivaldo J. Tro
ISBN: 9780321809247

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