Chapter 21, Problem 21.116QP

(a)

Interpretation Introduction

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.

Answer to Problem 21.116QP

The complete and balanced equation for the given equation is:

${\text{3Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + 2Al(s)}\stackrel{}{\to }{\text{ 3Pb(s) + 2Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

Explanation of Solution

Given Equation:

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + Al(s)}\stackrel{}{\to }$

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,

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + Al(s)}\stackrel{}{\to }{\text{ Pb(s) + Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

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.

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + Al(s)}\stackrel{}{\to }{\text{ Pb(s) + Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

Reactant side	Atom	Product side
1	$\text{Al}$	1
2	${\text{(NO}}_{\text{3}})$	3
1	$\text{Pb}$	1

To balance the above equation, multiply ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{ and Pb}$ by three and ${\text{Al(s) and Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}$ by two

${\text{3Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + 2Al(s)}\stackrel{}{\to }{\text{ 3Pb(s) + 2Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

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	$\text{Al}$	2
6	${\text{(NO}}_{\text{3}})$	6
3	$\text{Pb}$	3

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:

  ${\text{3Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + 2Al(s)}\stackrel{}{\to }{\text{ 3Pb(s) + 2Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

Conclusion

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

${\text{3Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}\text{(aq) + 2Al(s)}\stackrel{}{\to }{\text{ 3Pb(s) + 2Al(NO}}_{\text{3}}{\text{)}}_{\text{3}}\text{(aq)}$

(b)

Interpretation Introduction

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.

Answer to Problem 21.116QP

The complete and balanced equation of the given equation is:

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + 2NaNO}}_{\text{3}}(aq)$

Explanation of Solution

Given Equation:

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }$

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,

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + NaNO}}_{\text{3}}(aq)$

Balancing the equation:

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

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + NaNO}}_{\text{3}}(aq)$

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	$\text{Pb}$	1
10	$\text{O}$	7
2	$\text{Na}$	1
2	$\text{N}$	1
1	$\text{Cr}$	1

To balance the above equation, multiply ${\text{NaNO}}_{\text{3}}$ on the product side by two.

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + 2NaNO}}_{\text{3}}(aq)$

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	$\text{Pb}$	1
10	$\text{O}$	10
2	$\text{Na}$	2
2	$\text{N}$	2
1	$\text{Cr}$	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:

  ${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + 2NaNO}}_{\text{3}}(aq)$

Conclusion

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

${\text{Pb(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{ + Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ PbCrO}}_{\text{4}}{\text{(s) + 2NaNO}}_{\text{3}}(aq)$

(c)

Interpretation Introduction

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.

Answer to Problem 21.116QP

The complete and balanced equation for the given equation is:

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + 6LiOH(aq)}\stackrel{}{\to }{\text{ 2Al(OH)}}_{\text{3}}{\text{(s) + 3Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Explanation of Solution

Given Equation:

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + dilute LiOH(aq)}\stackrel{}{\to }$

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,

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + LiOH(aq)}\stackrel{}{\to }{\text{ Al(OH)}}_{\text{3}}{\text{(s) + Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

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.

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + LiOH(aq)}\stackrel{}{\to }{\text{ Al(OH)}}_{\text{3}}{\text{(s) + Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Reactant side	Atom	Product side
2	$\text{Al}$	1
13	$\text{O}$	7
1	$\text{H}$	3
3	$\text{S}$	1
1	$\text{Li}$	2

To balance the above equation, multiply $\text{LiOH}$ on the reactant side by six and ${\text{Al(OH)}}_{\text{3}}$ on the product side by two and ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ by three

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + 6LiOH(aq)}\stackrel{}{\to }{\text{ 2Al(OH)}}_{\text{3}}{\text{(s) + 3Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

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	$\text{Al}$	2
18	$\text{O}$	18
3	$\text{H}$	6
3	$\text{S}$	3
6	$\text{Li}$	6

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:

  ${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + 6LiOH(aq)}\stackrel{}{\to }{\text{ 2Al(OH)}}_{\text{3}}{\text{(s) + 3Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

Conclusion

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

${\text{Al}}_{\text{2}}{{\text{(SO}}_4\text{)}}_3\text{(aq) + 6LiOH(aq)}\stackrel{}{\to }{\text{ 2Al(OH)}}_{\text{3}}{\text{(s) + 3Li}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}$

(d)

Interpretation Introduction

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.

Answer to Problem 21.116QP

The complete and balanced equation for the given equation is:

$\text{2Al(s) + 6HCl(aq)}\stackrel{}{\to }{\text{ 3H}}_{\text{2}}{\text{(g) + 2AlCl}}_{\text{3}}\text{(aq)}$

Explanation of Solution

Given Equation:

$\text{Al(s) + HCl(aq)}\stackrel{}{\to }$

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,

$\text{Al(s) + HCl(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + AlCl}}_{\text{3}}\text{(aq)}$

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.

$\text{Al(s) + HCl(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + AlCl}}_{\text{3}}\text{(aq)}$

Reactant side	Atom	Product side
1	$\text{Al}$	1
1	$\text{H}$	2
1	$\text{Cl}$	3

Multiply $\text{Al and HCl}$ on the reactant side by two and six respectively, then multiply ${\text{H}}_{\text{2}}{\text{ and AlCl}}_{\text{3}}$ on the product side by three and two respectively.

$\text{2Al(s) + 6HCl(aq)}\stackrel{}{\to }{\text{ 3H}}_{\text{2}}{\text{(g) + 2AlCl}}_{\text{3}}\text{(aq)}$

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	$\text{Al}$	2
6	$\text{H}$	6
6	$\text{Cl}$	6

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:

  $\text{2Al(s) + 6HCl(aq)}\stackrel{}{\to }{\text{ 3H}}_{\text{2}}{\text{(g) + 2AlCl}}_{\text{3}}\text{(aq)}$

Conclusion

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

$\text{2Al(s) + 6HCl(aq)}\stackrel{}{\to }{\text{ 3H}}_{\text{2}}{\text{(g) + 2AlCl}}_{\text{3}}\text{(aq)}$

(e)

Interpretation Introduction

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.

Answer to Problem 21.116QP

The complete and balanced equation for the given equation is:

$\text{Sn(s) + 2HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$

Explanation of Solution

Given Equation:

$\text{Sn(s) + HBr(aq)}\stackrel{}{\to }$

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,

$\text{Sn(s) + HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$

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.

$\text{Sn(s) + HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$

Reactant side	Atom	Product side
1	$\text{Sn}$	1
1	$\text{Br}$	2
1	$\text{H}$	2

Multiply $\text{HBr}$ on the reactant side by two.

$\text{Sn(s) + 2HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$

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	$\text{Sn}$	1
2	$\text{Br}$	2
2	$\text{H}$	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:

  $\text{Sn(s) + 2HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$

Conclusion

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

$\text{Sn(s) + 2HBr(aq)}\stackrel{}{\to }{\text{ H}}_{\text{2}}{\text{(g) + SnBr}}_{\text{2}}\text{(aq)}$