Student Solutions Manual for Ebbing/Gammon's General Chemistry
Student Solutions Manual for Ebbing/Gammon's General Chemistry
11th Edition
ISBN: 9781305886780
Author: Darrell Ebbing; Steven D. Gammon
Publisher: Cengage Learning US
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Chapter 21, Problem 21.104QP
Interpretation Introduction

(a)

Interpretation:

The given equation has to be written complete and balanced

Concept Introduction:

Balancing the equation:

  • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.
  • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
  • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

To Write: The given equation complete and balanced.

Expert Solution
Check Mark

Answer to Problem 21.104QP

The complete and balanced equation for the given equation is:

2KOH(aq) + MgCl2(aq) Mg(OH)2(s) + 2KCl(aq)

Explanation of Solution

Given Equation:

KOH(aq) + MgCl2(aq) 

Complete Equation:

A complete equation will have same elements present on both sides of the equation.

The above equation can be completed by writing as follows,

KOH(aq) + MgCl2(aq) Mg(OH)2(s) + KCl(aq)

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

List the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

KOH(aq) + MgCl2(aq) Mg(OH)2(s) + KCl(aq)

Reactant side Atom Product side
1 K 1
1 O 2
1 H 2
1 Mg 1
2 Cl 1

To balance the Cl, H and O atom, multiply KOH on the reactant side and KCl on the product side by two.

2KOH(aq) + MgCl2(aq) Mg(OH)2(s) + 2KCl(aq)

Again, list the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Reactant side Atom Product side
2 K 2
2 O 2
2 H 2
1 Mg 1
2 Cl 2

The number of atoms on both sides of the reaction are same. Therefore, the above equation is a balanced one.

Hence, the complete and balanced form of the given equation is written as:

2KOH(aq) + MgCl2(aq) Mg(OH)2(s) + 2KCl(aq)

Conclusion

The complete and balanced equation of the given equation is written as:

2KOH(aq) + MgCl2(aq) Mg(OH)2(s) + 2KCl(aq)

Interpretation Introduction

(b)

Interpretation:

The given equation has to be written complete and balanced

Concept Introduction:

Balancing the equation:

  • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.
  • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
  • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

To Write: The given equation complete and balanced.

Expert Solution
Check Mark

Answer to Problem 21.104QP

The complete and balanced equation of the given equation is:

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

Explanation of Solution

Given Equation:

Mg(s) + CuSO4(aq) 

Complete equation:

A complete equation will have same present on both sides of the equation.

The above equation can be completed by writing as follows,

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

List the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Reactant side Atom Product side
1 Cu 1
4 O 4
1 Mg 1
1 S 1

The number of atoms on both sides of the reaction are same. Therefore, the above equation is a balanced one.

Hence, the complete and balanced form of the given equation is written as:

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

Conclusion

The complete and balanced equation of the given equation is written as:

Mg(s) + CuSO4(aq) Cu(s) + MgSO4(aq)

Interpretation Introduction

(c)

Interpretation:

The given equation has to be written complete and balanced

Concept Introduction:

Balancing the equation:

  • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.
  • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
  • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

To Write: The given equation complete and balanced.

Expert Solution
Check Mark

Answer to Problem 21.104QP

The complete and balanced equation for the given equation is:

Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g)

Explanation of Solution

Given Equation:

Sr(s) + H2O(l)

Complete Equation:

A complete equation will have same elements present on both sides of the equation.

The above equation can be completed by writing as follows,

Sr(s) + H2O(l) Sr(OH)2(aq) + H2(g)

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

List the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Sr(s) + H2O(l) Sr(OH)2(aq) + H2(g)

Reactant side Atom Product side
1 Sr 1
1 O 2
2 H 4

To balance the H and O atom, multiply water on the reactant side by two.

Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g)

Again, list the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Reactant side Atom Product side
1 Sr 1
2 O 2
4 H 4

The number of atoms on both sides of the reaction are same. Therefore, the above equation is a balanced one.

Hence, the complete and balanced form of the given equation is written as:

Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g)

Conclusion

The complete and balanced equation of the given equation is written as:

Sr(s) + 2H2O(l) Sr(OH)2(aq) + H2(g)

Interpretation Introduction

(d)

Interpretation:

The given equation has to be written complete and balanced

Concept Introduction:

Balancing the equation:

  • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.
  • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
  • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

To Write: The given equation complete and balanced.

Expert Solution
Check Mark

Answer to Problem 21.104QP

The complete and balanced equation for the given equation is:

SrCO3(s) + 2HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Explanation of Solution

Given Equation:

SrCO3(s) + HCl(aq)

Complete Equation:

A complete equation will have same elements present on both sides of the equation.

The above equation can be completed by writing as follows,

SrCO3(s) + HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

List the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

SrCO3(s) + HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Reactant side Atom Product side
1 Sr 1
3 O 3
1 H 2
1 Cl 2
1 C 1

To balance the Cl and H atoms, multiply HCl on the reactant side by two.

SrCO3(s) + 2HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Again, list the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Reactant side Atom Product side
1 Sr 1
3 O 3
2 H 2
2 Cl 2
1 C 1

The number of atoms on both sides of the reaction are same. Therefore, the above equation is a balanced one.

Hence, the complete and balanced form of the given equation is written as:

SrCO3(s) + 2HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Conclusion

The complete and balanced equation of the given equation is written as:

SrCO3(s) + 2HCl(aq) CO2(g) + H2O(l) + SrCl2(aq)

Interpretation Introduction

(e)

Interpretation:

The given equation has to be written complete and balanced

Concept Introduction:

Balancing the equation:

  • There is a Law for conversion of mass in a chemical reaction i.e., the mass of total amount of the product should be equal to the total mass of the reactants.
  • The concept of writing a balanced chemical reaction is depends on conversion of reactants into products.
  • First write the reaction from the given information.
  • Then count the number of atoms of each element in reactants as well as products.
  • Finally obtained values could place it as coefficients of reactants as well as products.
  • Loss of electron and loss of Hydrogen in a compound is oxidation - the compound is oxidized.  Gain of electron, gain of Oxygen in a compound is reduction - the compound is reduced.
  • Oxidation reduction and reduction reaction occur simultaneously in same reaction.

To Write: The given equation complete and balanced.

Expert Solution
Check Mark

Answer to Problem 21.104QP

The complete and balanced equation for the given equation is:

Ba(OH)2(aq) + CO2(aq) BaCO3(s) + H2O(l)

Explanation of Solution

Given Equation:

Ba(OH)2(aq) + CO2(aq)

Complete Equation:

A complete equation will have same elements present on both sides of the equation.

The above equation can be completed by writing as follows,

Ba(OH)2(aq) + CO2(aq) BaCO3(s) + H2O(l)

Balancing the equation:

A balanced equation will have same elements and same number of atoms of each side of the reaction.

List the atoms of the equation in a table and check for the equal number of atoms present on either side of the reaction.

Ba(OH)2(aq) + CO2(aq) BaCO3(s) + H2O(l)

Reactant side Atom Product side
1 Ba 1
4 O 4
2 H 2
1 C 1

The number of atoms on both sides of the reaction are same. Therefore, the above equation is a balanced one.

Hence, the complete and balanced form of the given equation is written as:

Ba(OH)2(aq) + CO2(aq) BaCO3(s) + H2O(l)

Conclusion

The complete and balanced equation of the given equation is written as:

Ba(OH)2(aq) + CO2(aq) BaCO3(s) + H2O(l)

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Chapter 21 Solutions

Student Solutions Manual for Ebbing/Gammon's General Chemistry

Ch. 21 - Prob. 21.9QPCh. 21 - Prob. 21.10QPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Prob. 21.40QPCh. 21 - Prob. 21.41QPCh. 21 - Describe the steps in the Ostwald process for the...Ch. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - What is the most important commercial means of...Ch. 21 - Prob. 21.50QPCh. 21 - Prob. 21.51QPCh. 21 - Prob. 21.52QPCh. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - Prob. 21.59QPCh. 21 - Prob. 21.60QPCh. 21 - Prob. 21.61QPCh. 21 - A test tube contains a solution of one of the...Ch. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - Prob. 21.68QPCh. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Prob. 21.71QPCh. 21 - Prob. 21.72QPCh. 21 - Prob. 21.73QPCh. 21 - Prob. 21.74QPCh. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - Prob. 21.77QPCh. 21 - Prob. 21.78QPCh. 21 - Prob. 21.79QPCh. 21 - Prob. 21.80QPCh. 21 - Prob. 21.81QPCh. 21 - Prob. 21.82QPCh. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85QPCh. 21 - Prob. 21.86QPCh. 21 - Sketch a diagram showing the formation of energy...Ch. 21 - Sketch a diagram showing the formation of energy...Ch. 21 - Prob. 21.89QPCh. 21 - Prob. 21.90QPCh. 21 - Prob. 21.91QPCh. 21 - Prob. 21.92QPCh. 21 - Prob. 21.93QPCh. 21 - Prob. 21.94QPCh. 21 - Francium was discovered as a minor decay product...Ch. 21 - Prob. 21.96QPCh. 21 - Prob. 21.97QPCh. 21 - Prob. 21.98QPCh. 21 - Prob. 21.99QPCh. 21 - Prob. 21.100QPCh. 21 - Prob. 21.101QPCh. 21 - Prob. 21.102QPCh. 21 - Prob. 21.103QPCh. 21 - Prob. 21.104QPCh. 21 - Prob. 21.105QPCh. 21 - Prob. 21.106QPCh. 21 - Prob. 21.107QPCh. 21 - Prob. 21.108QPCh. 21 - Prob. 21.109QPCh. 21 - Prob. 21.110QPCh. 21 - Prob. 21.111QPCh. 21 - Prob. 21.112QPCh. 21 - Prob. 21.113QPCh. 21 - Prob. 21.114QPCh. 21 - Prob. 21.115QPCh. 21 - Prob. 21.116QPCh. 21 - Prob. 21.117QPCh. 21 - Prob. 21.118QPCh. 21 - Prob. 21.119QPCh. 21 - Prob. 21.120QPCh. 21 - Prob. 21.121QPCh. 21 - Prob. 21.122QPCh. 21 - Prob. 21.123QPCh. 21 - Prob. 21.124QPCh. 21 - Prob. 21.125QPCh. 21 - Prob. 21.126QPCh. 21 - Prob. 21.127QPCh. 21 - Prob. 21.128QPCh. 21 - Prob. 21.129QPCh. 21 - Prob. 21.130QPCh. 21 - Prob. 21.131QPCh. 21 - Prob. 21.132QPCh. 21 - Prob. 21.133QPCh. 21 - Prob. 21.134QPCh. 21 - Prob. 21.135QPCh. 21 - Prob. 21.136QPCh. 21 - Prob. 21.137QPCh. 21 - Prob. 21.138QPCh. 21 - Prob. 21.139QPCh. 21 - Prob. 21.140QPCh. 21 - Prob. 21.141QPCh. 21 - Prob. 21.142QPCh. 21 - Prob. 21.143QPCh. 21 - Phosphorous acid, H3PO3, is oxidized to phosphoric...Ch. 21 - Prob. 21.145QPCh. 21 - Prob. 21.146QPCh. 21 - Prob. 21.147QPCh. 21 - Prob. 21.148QPCh. 21 - What are the oxidation numbers of sulfur in each...Ch. 21 - What are the oxidation numbers of sulfur in each...Ch. 21 - Prob. 21.151QPCh. 21 - Prob. 21.152QPCh. 21 - Prob. 21.153QPCh. 21 - Prob. 21.154QPCh. 21 - Prob. 21.155QPCh. 21 - Prob. 21.156QPCh. 21 - Chlorine can be prepared by oxidizing chloride ion...Ch. 21 - Prob. 21.158QPCh. 21 - Prob. 21.159QPCh. 21 - Prob. 21.160QPCh. 21 - Prob. 21.161QPCh. 21 - Prob. 21.162QPCh. 21 - Prob. 21.163QPCh. 21 - Prob. 21.164QPCh. 21 - Prob. 21.165QPCh. 21 - Prob. 21.166QPCh. 21 - Prob. 21.167QPCh. 21 - Xenon trioxide, XeO3, is reduced to xenon in...Ch. 21 - Prob. 21.169QPCh. 21 - Prob. 21.170QPCh. 21 - Prob. 21.171QPCh. 21 - Prob. 21.172QPCh. 21 - Prob. 21.173QPCh. 21 - Prob. 21.174QPCh. 21 - Prob. 21.175QPCh. 21 - Prob. 21.176QPCh. 21 - Prob. 21.177QPCh. 21 - Prob. 21.178QPCh. 21 - Prob. 21.179QPCh. 21 - Prob. 21.180QPCh. 21 - Prob. 21.181QPCh. 21 - Prob. 21.182QPCh. 21 - Prob. 21.183QPCh. 21 - Prob. 21.184QPCh. 21 - Prob. 21.185QPCh. 21 - Prob. 21.186QPCh. 21 - Prob. 21.187QPCh. 21 - Sodium perchlorate, NaClO4, is produced by...Ch. 21 - The amount of sodium hypochlorite in a bleach...Ch. 21 - Prob. 21.190QPCh. 21 - Prob. 21.191QPCh. 21 - Prob. 21.192QPCh. 21 - Prob. 21.193QPCh. 21 - Prob. 21.194QPCh. 21 - Prob. 21.195QPCh. 21 - Prob. 21.196QPCh. 21 - Prob. 21.197QPCh. 21 - Prob. 21.198QPCh. 21 - Prob. 21.199QPCh. 21 - Prob. 21.200QPCh. 21 - Prob. 21.201QPCh. 21 - Prob. 21.202QPCh. 21 - Prob. 21.203QPCh. 21 - Prob. 21.204QP
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