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A 55.0 mL sample of a 0.102 M potassium sulfate solution

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

Solution for problem 70E 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 70E

A 55.0 mL sample of a 0.102 M potassium sulfate solution is mixed with 35.0 mL of a 0.114 M lead(II) acetate solution and this precipitation reaction occurs:

The solid PbSO4 is collected, dried, and found to have a mass of 1.01 g. Determine the limiting reactant, the theoretical yield, and the percent yield.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to identify which of the given pairs of solution has higher concentration of iodide ion. Step1: a) 1 molecule of BaI dissociates in aqueous solution to give 1 Ba ion and 2 I ions. 2+ - 2 - Therefore, 0.10 M BaI will g2e (2 x 0.10 = 0.20) M I ions in solution. Similarly, 1 molecule of KI dissociates in aqueous solution to give one K ion and one I ion. + - - Therefore, 0.25 M KI will produce 0.25 M I ions in solution. - Thus, 0.25 M KI solution has higher concentration of I ions. Step2: b) Number of moles of any compound in solution is given by: Number of moles of solute = Volume of solution in L X Molarity solute Therefore, number of moles of KI = (100 / 1000 L) X 0.10 M = 0.1 L X 0.10 M = 0.01 moles Similarly, number of moles of ZnI = (200 / 1000 L) X 0.040 M 2 = 0.2 L X 0.040 M =0.008...

Step 2 of 3

Chapter 4, Problem 70E is Solved
Step 3 of 3

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

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A 55.0 mL sample of a 0.102 M potassium sulfate solution

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