General Chemistry
General Chemistry
4th Edition
ISBN: 9781891389603
Author: Donald A. McQuarrie, Peter A. Rock, Ethan B. Gallogly
Publisher: University Science Books
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 11, Problem 11.56P

a)

Interpretation Introduction

Interpretation:

Balanced chemical equation for reaction of cadmium chloride with silver perchlorate has to be determined.

Concept Introduction:

The reaction that has equal number of atoms of different elements in reactant as well as in product side is called balanced chemical reaction. The chemical equation is balanced to follow the Law of the conservation of mass.

a)

Expert Solution
Check Mark

Explanation of Solution

The steps to balance a chemical reaction are as follows:

Step 1: Write the unbalanced equation.

  CdCl2+AgClO4AgCl+Cd(ClO4)2

Step 2:Then write the number of atoms of all elements that are present in chemical reaction in the reactant side and product side.

i) On reactant side,

Number of silver atom is 1.

Number of cadmium atom is 1.

Number of oxygen atoms is 4.

Number of chlorine atom is 3.

ii) On product side,

Number of silver atom is 1.

Number of cadmium atom is 1.

Number of oxygen atoms is 8.

Number of chlorine atom is 3.

Step 3: Balance the number of other atoms of elements except oxygen. The other atom are balanced on both side. Now the reaction is,

  CdCl2+AgClO4AgCl+Cd(ClO4)2

Step 4: After this, balance the number of oxygen atoms. The oxygen atoms are unbalanced on both sides. Multiply AgClO4 by 2 to balance oxygen atom and AgCl by 2 to balance silver atom as silver atom will become unbalanced. Now the reaction is,

  CdCl2+2AgClO42AgCl+CdClO4

Step 5: Finally, check the number of atoms of each element on both sides. If the number is same then chemical equation is balanced. The balanced chemical equation is,

  CdCl2+2AgClO42AgCl+Cd(ClO4)2

b)

Interpretation Introduction

Interpretation:

Mass of AgCl produced from the reaction has to be determined.

Concept Introduction:

Mole is S.I. unit. The number of moles is calculated as ratio of mass of compound to molar mass of compound.

Molar mass is sum of the total mass in grams of all atoms that make up mole of particular molecule that is mass of 1 mole of compound. The S.I unit is g/mol.

The expression to relate number of moles, mass and molar mass of compound is as follows:

  Number of moles=mass of the compoundmolar mass of the compound

For more than one quantity of reactants in a reaction, the reactant that is completely consumed and controls the amount of product synthesized is called limiting reactant. The other reactants are called excess reactants. Also, the limiting reactant is completely consumed the initial mass of limiting reactant is used to calculate mass of product.

b)

Expert Solution
Check Mark

Answer to Problem 11.56P

Mass of AgCl produced from the reaction is 11.47108 g.

Explanation of Solution

The reaction is as follows:

    CdCl2+2AgClO42AgCl+Cd(ClO4)2

The formula to calculate moles of CdCl2 is as follows:

  Number of moles of CdCl2=mass of CdCl2molar mass of CdCl2        (1)

Substitute 17.5 g for mass of CdCl2 and 183.317 g/mol for molar mass of CdCl2 in equation (1).

  Number of moles of CdCl2=(17.5 g183.317 g/mol)=0.09546 mol

The formula to calculate moles of AgClO4 is as follows:

  Number of moles ofAgClO4=mass ofAgClO4molar massof AgClO4         (2)

Substitute 35.5 g for mass of AgClO4 and 207.3188 g/mol for molar mass of AgClO4 in equation (2).

  Number of moles ofAgClO4=(35.5 g207.3188 g/mol)=0.1712 mol

To find the limiting reactant, divide the number of moles of each reactant by their stoichiometric coefficients.

i) The stoichiometric coefficient of CdCl2 is 1 and number of moles of CdCl2 is 0.09546 mol. Therefore, number of molesof CdCl2 will be divided by 1. Hence, number of moles CdCl2 will be 0.09546 mol.

ii)The stoichiometric coefficient of AgClO4 is 2 and number of moles of AgClO4 is 0.1712 mol. Therefore, number of moles of AgClO4 will be divided by 2. Hence, number of moles AgClO4 will be 0.0856 mol.

The number of moles of AgClO4 is less in comparison to other reactants in reaction. Hence, AgClO4 is limiting reagent. Also, initial mass of AgClO4 is used to calculate mass of product formed.

The reaction between CdCl2 and AgClO4 is as follows:

    CdCl2+2AgClO42AgCl+Cd(ClO4)2

According to stoichiometry of reaction, 2 moles of AgClO4 produces 2 moles of AgCl. Therefore, number of moles of AgCl is equal to number of moles of AgClO4 and that is 0.1712 mol.

The formula to calculate mass of AgCl is as follows:

    Mass of AgCl=(number of moles)(molar mass)        (3)

Substitute 143.3212 g/mol for molar mass of AgCl and 0.1712 mol for number of moles of AgCl in equation (3).

  Mass of AgCl=(0.1712 mol)(143.3212 g/mol)=24.5365 g

c)

Interpretation Introduction

Interpretation:

Mass of excess reactant among CdCl2 and AgClO4 has to be determined.

Concept Introduction:

Refer to part (b).

c)

Expert Solution
Check Mark

Answer to Problem 11.56P

Mass of excess CdCl2 is 1.8075 g.

Explanation of Solution

The reaction between CdCl2 and AgClO4 is as follows:

    CdCl2+2AgClO42AgCl+Cd(ClO4)2

The number of moles of AgClO4 is less in comparison to other reactants in reaction. Hence, AgClO4 is limiting reagent and CdCl2 will be excess reagent.

According to stoichiometry of reaction, 2 moles of AgClO4 reacts with 1 mole of CdCl2. Therefore, number of moles of CdCl2 is 1/2 times number of moles of AgClO4.

The formula to calculate number of moles of CdCl2 is as follows:

    Number of moles of CdCl2=12(Number of moles of AgClO4)        (4)

Substitute 0.1712 mol for number moles of AgClO4 in equation (4).

  Number of moles of CdCl2=12(0.1712 mol)=0.0856mol

Excess mole of CdCl2 is calculated as follows:

    Excess mole of CdCl2=moles of CdCl2mole of CdCl2 required        (5)

Substitute, 0.09546 mol for moles of CdCl2 and 0.0856 mol for moles of CdCl2 require in equation (5).

  Excess mole of CdCl2=0.09546 mol0.0856 mol=0.00986 mol

The formula to calculate mass of CdCl2 in excess is as follows:

    Mass of excess of CdCl2=(number of moles in excess)(molar mass)        (6)

Substitute 0.00986 mol for excess moles of CdCl2 and 183.317 g/mol for molar mass of CdCl2 in equation (6).

  Mass of excess of CdCl2=(0.00986 mol)(183.317 g/mol)=1.8075 g

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 11 Solutions

General Chemistry

Ch. 11 - Prob. 11.11PCh. 11 - Prob. 11.12PCh. 11 - Prob. 11.13PCh. 11 - Prob. 11.14PCh. 11 - Prob. 11.15PCh. 11 - Prob. 11.16PCh. 11 - Prob. 11.17PCh. 11 - Prob. 11.18PCh. 11 - Prob. 11.19PCh. 11 - Prob. 11.20PCh. 11 - Prob. 11.21PCh. 11 - Prob. 11.22PCh. 11 - Prob. 11.23PCh. 11 - Prob. 11.24PCh. 11 - Prob. 11.25PCh. 11 - Prob. 11.26PCh. 11 - Prob. 11.27PCh. 11 - Prob. 11.28PCh. 11 - Prob. 11.29PCh. 11 - Prob. 11.30PCh. 11 - Prob. 11.31PCh. 11 - Prob. 11.32PCh. 11 - Prob. 11.33PCh. 11 - Prob. 11.34PCh. 11 - Prob. 11.35PCh. 11 - Prob. 11.36PCh. 11 - Prob. 11.37PCh. 11 - Prob. 11.38PCh. 11 - Prob. 11.39PCh. 11 - Prob. 11.40PCh. 11 - Prob. 11.41PCh. 11 - Prob. 11.42PCh. 11 - Prob. 11.43PCh. 11 - Prob. 11.44PCh. 11 - Prob. 11.45PCh. 11 - Prob. 11.46PCh. 11 - Prob. 11.47PCh. 11 - Prob. 11.48PCh. 11 - Prob. 11.49PCh. 11 - Prob. 11.50PCh. 11 - Prob. 11.51PCh. 11 - Prob. 11.52PCh. 11 - Prob. 11.53PCh. 11 - Prob. 11.54PCh. 11 - Prob. 11.55PCh. 11 - Prob. 11.56PCh. 11 - Prob. 11.57PCh. 11 - Prob. 11.58PCh. 11 - Prob. 11.59PCh. 11 - Prob. 11.60PCh. 11 - Prob. 11.61PCh. 11 - Prob. 11.62PCh. 11 - Prob. 11.63PCh. 11 - Prob. 11.64PCh. 11 - Prob. 11.65PCh. 11 - Prob. 11.66PCh. 11 - Prob. 11.67PCh. 11 - Prob. 11.68PCh. 11 - Prob. 11.69PCh. 11 - Prob. 11.70PCh. 11 - Prob. 11.71PCh. 11 - Prob. 11.72PCh. 11 - Prob. 11.73PCh. 11 - Prob. 11.74PCh. 11 - Prob. 11.75PCh. 11 - Prob. 11.76PCh. 11 - Prob. 11.77PCh. 11 - Prob. 11.78PCh. 11 - Prob. 11.79PCh. 11 - Prob. 11.80PCh. 11 - Prob. 11.81PCh. 11 - Prob. 11.82PCh. 11 - Prob. 11.83PCh. 11 - Prob. 11.84PCh. 11 - Prob. 11.85PCh. 11 - Prob. 11.86PCh. 11 - Prob. 11.87PCh. 11 - Prob. 11.88PCh. 11 - Prob. 11.89PCh. 11 - Prob. 11.90PCh. 11 - Prob. 11.91PCh. 11 - Prob. 11.92PCh. 11 - Prob. 11.93P
Knowledge Booster
Background pattern image
Chemistry
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Text book image
Chemistry
Chemistry
ISBN:9781259911156
Author:Raymond Chang Dr., Jason Overby Professor
Publisher:McGraw-Hill Education
Text book image
Principles of Instrumental Analysis
Chemistry
ISBN:9781305577213
Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:Cengage Learning
Text book image
Organic Chemistry
Chemistry
ISBN:9780078021558
Author:Janice Gorzynski Smith Dr.
Publisher:McGraw-Hill Education
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
Elementary Principles of Chemical Processes, Bind...
Chemistry
ISBN:9781118431221
Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:WILEY
Bonding (Ionic, Covalent & Metallic) - GCSE Chemistry; Author: Science Shorts;https://www.youtube.com/watch?v=p9MA6Od-zBA;License: Standard YouTube License, CC-BY
Stoichiometry - Chemistry for Massive Creatures: Crash Course Chemistry #6; Author: Crash Course;https://www.youtube.com/watch?v=UL1jmJaUkaQ;License: Standard YouTube License, CC-BY