   Chapter 9, Problem 73AP ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425

#### Solutions

Chapter
Section ### Introductory Chemistry: A Foundati...

9th Edition
Steven S. Zumdahl + 1 other
ISBN: 9781337399425
Textbook Problem
3 views

# The traditional method of analysis for the amount of chloride ion present in a sample is to dissolve the sample in water and then slowly to add a solution of silver nitrate. Silver chloride ¡s very insoluble in water, and by adding a slight excess of silver nitrate, it is possible to effectively remove all chloride ion from the sample.:math> Ag + ( a q ) + Cl + ( a q ) → AgCl ( s ) ppose a 1.054-g sample is known to contain 10.3% chloride ion by mass. What mass of silver nitrate must be used to completely precipitate the chloride ion from the sample? What mass of silver chloride will be obtained?

Interpretation Introduction

Interpretation:

Mass of silver nitrate must be used to completely precipitate the chloride ion from the 1.054 g sample which contain 10.3 % chloride ion by mass should be calculated. Mass of silver chloride obtained should be calculated.

Concept Introduction:

Moles represent numbers of molecules, and we cannot count molecules directly. In chemistry, we count by weighing. The process of using chemical equation to calculate the relative masses of reactants and products involved in a reaction is called stoichiometry. The balanced equation for a chemical reaction describes the stoichiometry of the reaction.

Explanation

Given information:

Moles represent numbers of molecules, and we cannot count molecules directly. In chemistry, we count by weighing. The process of using chemical equation to calculate the relative masses of reactants and products involved in a reaction is called stoichiometry. The balanced equation for a chemical reaction describes the stoichiometry of the reaction.

Balanced equation for the reaction is,

AgNO3(aq)+ Cl(aq)     AgCl(s)+ NO3(aq)

Mass of chloride ions present in the sample = 1.054 g × 10.3100

= 0.109 g

Molar mass of Cl = 35.45 g/mol

Molar mass=Mass of the sample in gramsNumber of moles

Number of moles=Mass of the sample in gramsMolar mass

Number of moles of chloride ions present in a mass of 0.109 g = 0.109 g35.45 g/mol

= 0.00307 mol

Mole ratio between AgNO3 and Cl- = 1 mol1 mol

Number of moles of AgNO3 needed to react completely with Cl- = 0

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