Chapter 21, Problem 21.115QP

(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 21.115QP

The complete and balanced equation for the given equation is:

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 3H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Explanation of Solution

Given Equation:

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }$

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,

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

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.

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + H}}_{\text{2}}\text{O(l)}$

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

To balance the above equation, multiply ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ on the reactant side and ${\text{H}}_{\text{2}}\text{O}$ on the product side by three.

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 3H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

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
15	$\text{O}$	15
6	$\text{H}$	6
3	$\text{S}$	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:

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 3H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

Conclusion

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

${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}{\text{(s) + 3H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{(aq)}\stackrel{}{\to }{\text{ Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_3{\text{(aq) + 3H}}_{\text{2}}\text{O(l)}$

(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 21.115QP

The complete and balanced equation of the given equation is:

${\text{Al(s) + 3AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ 3Ag(s) + Al(NO}}_3{)}_3(aq)$

Explanation of Solution

Given Equation:

${\text{Al(s) + AgNO}}_3\text{(aq)}\stackrel{}{\to }$

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,

${\text{Al(s) + AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ Ag(s) + Al(NO}}_3{)}_3(aq)$

Balancing the equation:

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

${\text{Al(s) + AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ Ag(s) + Al(NO}}_3{)}_3(aq)$

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	$\text{Al}$	1
3	$\text{O}$	9
1	$\text{Ag}$	1
1	$\text{N}$	3

To balance the above equation, multiply ${\text{AgNO}}_{\text{3}}$ on the reactant side and $\text{Ag}$ on the product side by three.

${\text{Al(s) + 3AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ 3Ag(s) + Al(NO}}_3{)}_3(aq)$

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	$\text{Al}$	1
9	$\text{O}$	9
3	$\text{Ag}$	3
3	$\text{N}$	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:

${\text{Al(s) + 3AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ 3Ag(s) + Al(NO}}_3{)}_3(aq)$

Conclusion

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

${\text{Al(s) + 3AgNO}}_3\text{(aq)}\stackrel{}{\to }{\text{ 3Ag(s) + Al(NO}}_3{)}_3(aq)$

(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 21.115QP

The complete and balanced equation for the given equation is:

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

Explanation of Solution

Given Equation:

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

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,

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

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.

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

Reactant side	Atom	Product side
1	$\text{Pb}$	1
6	$\text{O}$	3
2	$\text{N}$	1
1	$\text{Na}$	1
1	$\text{I}$	2

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

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

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
1	$\text{Pb}$	1
6	$\text{O}$	6
2	$\text{N}$	2
2	$\text{Na}$	2
2	$\text{I}$	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:

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

Conclusion

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

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

(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 21.115QP

The complete and balanced equation for the given equation is:

${\text{8Al(s) + 3Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ 9Mn(s) + 4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

Explanation of Solution

Given Equation:

${\text{Al(s) + Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }$

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,

${\text{Al(s) + Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ Mn(s) + Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

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.

${\text{Al(s) + Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ Mn(s) + Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

Reactant side	Atom	Product side
1	$\text{Al}$	2
4	$\text{O}$	3
3	$\text{Mn}$	1

Multiply ${\text{Mn}}_{\text{3}}{\text{O}}_{\text{4}}$ on the reactant side by three and multiply ${\text{Al}}_{\text{2}}{\text{O}}_{\text{3}}$ on the product side by four.

${\text{Al(s) + 3Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ Mn(s) + 4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

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	$\text{Al}$	8
12	$\text{O}$	12
9	$\text{Mn}$	1

Multiply $\text{Al}$ atom on the reactant side by eight and $\text{Mn}$ atom on the product side by nine and check the number of atoms again.

${\text{8Al(s) + 3Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ 9Mn(s) + 4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

Reactant side	Atom	Product side
8	$\text{Al}$	8
12	$\text{O}$	12
9	$\text{Mn}$	9

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:

${\text{8Al(s) + 3Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ 9Mn(s) + 4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

Conclusion

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

${\text{8Al(s) + 3Mn}}_{\text{3}}{\text{O}}_{\text{4}}\text{(s)}\stackrel{}{\to }{\text{ 9Mn(s) + 4Al}}_{\text{2}}{\text{O}}_{\text{3}}\text{(s)}$

(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 21.115QP

The complete and balanced equation for the given equation is:

${\text{2Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{O(g)}$

Explanation of Solution

Given Equation:

${\text{Ga(OH)}}_3\text{(s) }\stackrel{}{\to }$

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,

${\text{Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + H}}_{\text{2}}\text{O(g)}$

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.

${\text{Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + H}}_{\text{2}}\text{O(g)}$

Reactant side	Atom	Product side
1	$\text{Ga}$	2
3	$\text{O}$	4
3	$\text{H}$	2

Multiply ${\text{Ga(OH)}}_3$ on the reactant side by two and multiply ${\text{H}}_{\text{2}}\text{O}$ on the product side by three.

${\text{2Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{O(g)}$

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
2	$\text{Ga}$	2
6	$\text{O}$	6
6	$\text{H}$	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:

${\text{2Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{O(g)}$

Conclusion

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

${\text{2Ga(OH)}}_3\text{(s) }\stackrel{}{\to }{\text{ Ga}}_{\text{2}}{\text{O}}_3{\text{(s) + 3H}}_{\text{2}}\text{O(g)}$