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Lead ions can be precipitated from solution with NaCl

Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro ISBN: 9780321910295 34

Solution for problem 65P Chapter 8

Introductory Chemistry | 5th Edition

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Introductory Chemistry | 5th Edition | ISBN: 9780321910295 | Authors: Nivaldo J Tro

Introductory Chemistry | 5th Edition

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Problem 65P

PROBLEM 65P

Lead ions can be precipitated from solution with NaCl according to the reaction:

Pb2+ (aq) + 2 NaCl(aq) → PbCl2(s) + 2 Na+(aq)

When 135.8 g of NaCl are added to a solution containing 195.7 g of Pb2+, a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 252.4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction.

Step-by-Step Solution:
Step 1 of 3

Solution 65P

In this question , the given mass of NaCl = 135.8 g

                                Mass of lead = 195.7g

                The mass of PbCl2 precipitate = 252.4g

Pb2++ 2 NaCl → PbCl2(s) + 2 Na+ 

Part A: 

The molecular mass of NaCl = 58.4430 g/mol and atomic weight of

Pb = 207.2100g/mol

Hence number of mol of NaCl

=(135.8 g NaCl) / (58.4430 g NaCl/mol) = 2.3236 mol NaCl

Number of mol of Pb

= (195.7 g Pb) / (207.2100 g Pb/mol) = 0.944452 mol Pb

0.944452 mole of Pb would react completely with 0.944452 (2/1)...

Step 2 of 3

Chapter 8, Problem 65P is Solved
Step 3 of 3

Textbook: Introductory Chemistry
Edition: 5
Author: Nivaldo J Tro
ISBN: 9780321910295

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