Lead ions can be precipitated from solution with KCl

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

Problem 49E 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 49E

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

When 28.5 g KCl is added to a solution containing 25.7 g Pb2+ , a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction.

Step-by-Step Solution:
Step 1 of 3

Solution: Here, we are going to determine the oxidation number of the indicated element in the given set of compounds. Step1: Oxidation number denotes the oxidation state of an element in a compound ascertained according to a set of rules. These rules are: 1. In elements, in the free or the uncombined state, each atom bears an oxidation number of zero. Evidently each atom in H2, O2, Cl2, O3, P4, S8, Na, Mg, Al has the oxidation number zero. 2. For ions composed of only one atom, the oxidation number is equal to the charge on the ion. In their compounds all alkali metals have oxidation number of +1, and all alkaline earth metals have an oxidation number of +2. Aluminium is regarded to have an oxidation number of +3 in all its compounds. 3. The oxidation number of oxygen in most compounds is –2. 4. The oxidation number of hydrogen is +1, except when it is bonded to metals in binary compounds (that is compounds containing two elements). 5. In all its compounds, fluorine has an oxidation number of –1. Other halogens (Cl, Br,and I) also have an oxidation number of –1, when they occur as halide ions in their compounds. 6. The algebraic sum of the oxidation number of all the atoms in a compound must be zero. In polyatomic ion, the algebraic sum of all the oxidation numbers of atoms of the ion must equal the charge on the ion. Step2: Based upon these rules, let us calculate the oxidation number of the indicated elements in the given compounds: a)...

Step 2 of 3

Chapter 4, Problem 49E is Solved
Step 3 of 3

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

This full solution covers the following key subjects: oxidation, kmno, element, hbro, indicated. This expansive textbook survival guide covers 82 chapters, and 9464 solutions. The answer to “Lead ions can be precipitated from solution with KCl according to the reaction: When 28.5 g KCl is added to a solution containing 25.7 g Pb2+ , a PbCl2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29.4 g. Determine the limiting reactant, theoretical yield of PbCl2, and percent yield for the reaction.” is broken down into a number of easy to follow steps, and 60 words. This textbook survival guide was created for the textbook: Chemistry: A Molecular Approach, edition: 3rd. The full step-by-step solution to problem: 49E from chapter: 4 was answered by , our top Chemistry solution expert on 02/22/17, 04:35PM. Chemistry: A Molecular Approach was written by and is associated to the ISBN: 9780321809247. Since the solution to 49E from 4 chapter was answered, more than 319 students have viewed the full step-by-step answer.

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