Chapter 3, Problem 3.60QP

(a)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (a) is ${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\to 2{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+O}}_{\text{2}}$

Explanation of Solution

To write balanced equation for ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+O}}_{\text{2}}$

• List the atoms of given equation and the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
2	N	2
5	O	6

• Place 1/2 on ${\text{O}}_{\text{2}}$ on the right hand side

${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}\to {\text{N}}_{\text{2}}{\text{O}}_{\text{4}}\text{+}\frac{1}{2}{\text{O}}_{\text{2}}$

Reactant side	Atom	Product side
2	N	2
5	O	5

• To make whole number multiply by two on both sides

${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\to 2{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+O}}_{\text{2}}$

Reactant side	Atom	Product side
4	N	4
10	O	10

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{2N}}_{\text{2}}{\text{O}}_{\text{5}}\to 2{\text{N}}_{\text{2}}{\text{O}}_{\text{4}}{\text{+O}}_{\text{2}}$

(b)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (b) is ${\text{2KNO}}_{\text{3}}\text{+}\to 2{\text{KNO}}_{\text{2}}+{\text{O}}_{\text{2}}$

Explanation of Solution

To write balanced equation for ${\text{KNO}}_{\text{3}}\text{+}\to {\text{KNO}}_{\text{2}}+{\text{O}}_{\text{2}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	K	1
1	N	1
3	O	4

• To tally the oxygen atom on both sides, place 1/2 for oxygen on right hand side and again balance the equation.

${\text{KNO}}_{\text{3}}\text{+}\to {\text{KNO}}_{\text{2}}+\frac{1}{2}{\text{O}}_{\text{2}}$

Reactant side	Atom	Product side
1	K	1
1	N	1
3	O	3

• To give the equation in simple whole number coefficients, multiply both sides by two.

${\text{2KNO}}_{\text{3}}\text{+}\to 2{\text{KNO}}_{\text{2}}+{\text{O}}_{\text{2}}$

Reactant side	Atom	Product side
2	K	2
2	N	2
6	O	6

Now, the atom count on both sides of the reaction is same, hence the equation is balanced. The balanced equation is written as ${\text{2KNO}}_{\text{3}}\text{+}\to 2{\text{KNO}}_{\text{2}}+{\text{O}}_{\text{2}}$

(c)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (c) is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O+2H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O+H}}_{\text{2}}\text{O}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
2	N	2
4	H	2
3	O	2

• Multiply two on water on the right hand side of the equaiton

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O+2H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	N	2
4	H	4
3	O	3

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}\to {\text{N}}_{\text{2}}{\text{O+2H}}_{\text{2}}\text{O}$

(d)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (d) is ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}\text{+}\to {\text{N}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}\text{+}\to {\text{N}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

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

• Multiply water by two on right hand side of the equation.

${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}\text{+}\to {\text{N}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	N	2
2	O	2
4	H	4

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}\text{+}\to {\text{N}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

(e)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (e) is ${\text{2NaHCO}}_{\text{3}}\to {\text{Na}}_{\text{2}}C{\text{O}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

Explanation of Solution

To write balanced equation for ${\text{NaHCO}}_{\text{3}}\to {\text{Na}}_{\text{2}}C{\text{O}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	Na	2
1	H	2
1	C	2
3	O	6

• Multiply by two on left hand side

${\text{2NaHCO}}_{\text{3}}\to {\text{Na}}_{\text{2}}C{\text{O}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

Reactant side	Atom	Product side
2	Na	2
2	H	2
2	C	2
6	O	6

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{2NaHCO}}_{\text{3}}\to {\text{Na}}_{\text{2}}C{\text{O}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

(f)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (f) is ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\to 6{\text{H}}_{\text{2}}\text{O}\text{+}4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Explanation of Solution

To write balanced equation for ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\to {\text{H}}_{\text{2}}\text{O}\text{+}{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
4	P	1
11	O	4
2	H	3

• To balance phosphorus place four on phosphoric acid

${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\to {\text{H}}_{\text{2}}\text{O}\text{+}4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Reactant side	Atom	Product side
4	P	1
11	O	16
2	H	12

• Place six on water to balance the equation further.

  ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\to 6{\text{H}}_{\text{2}}\text{O}\text{+}4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

Reactant side	Atom	Product side
4	P	4
16	O	16
12	H	12

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\to 6{\text{H}}_{\text{2}}\text{O}\text{+}4{\text{H}}_{\text{3}}{\text{PO}}_{\text{4}}$

(g)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (g) is ${\text{2HCl+CaCO}}_{\text{3}}\to {\text{CaCl}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

Explanation of Solution

To write balanced equation for ${\text{HCl+CaCO}}_{\text{3}}\to {\text{CaCl}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

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

• Multiply hydrogen chloride by two on reactant side.

${\text{2HCl+CaCO}}_{\text{3}}\to {\text{CaCl}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

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

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{2HCl+CaCO}}_{\text{3}}\to {\text{CaCl}}_{\text{2}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

(h)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (h) is ${\text{2Al+3H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{+3H}}_{\text{2}}$

Explanation of Solution

To write balanced equation for ${\text{Al+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{+H}}_{\text{2}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	Al	2
1	${\text{SO}}_{\text{4}}$	3
2	H	2

• To tally the Aluminium atom, multiply two on left hand side.

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

Reactant side	Atom	Product side
2	Al	2
1	${\text{SO}}_{\text{4}}$	3
2	H	2

• To balance the ${\text{SO}}_{\text{4}}$ units, mulitply sulphuric acid by three on reactant side

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

Reactant side	Atom	Product side
2	Al	2
3	${\text{SO}}_{\text{4}}$	3
6	H	2

• To balance finally, multiply hydrogen by three on product side.

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

Reactant side	Atom	Product side
2	Al	2
3	${\text{SO}}_{\text{4}}$	3
6	H	6

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{2Al+3H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}{\text{+3H}}_{\text{2}}$

(i)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (i) is ${\text{CO}}_{\text{2}}\text{+2KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{CO}}_{\text{2}}\text{+KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	K	2
3	O	4
1	C	1
1	H	2

• To tally the Potassium and hydrogen atom, multiply potassium hydroxide by two on left hand side.

${\text{CO}}_{\text{2}}\text{+2KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	K	2
3	O	3
1	C	1
2	H	2

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{CO}}_{\text{2}}\text{+2KOH}\to {\text{K}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}$

(j)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (j) is ${\text{CH}}_{\text{4}}{\text{+2O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{CH}}_{\text{4}}{\text{+O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	C	1
2	O	3
4	H	2

• Multiply water by two on right hand side

${\text{CH}}_{\text{4}}{\text{+O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

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

• Multiply oxygen atom by two on reactant side

${\text{CH}}_{\text{4}}{\text{+2O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

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

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{CH}}_{\text{4}}{\text{+2O}}_{\text{2}}\to {\text{CO}}_{\text{2}}+2{\text{H}}_{\text{2}}\text{O}$

(k)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (k) is ${\text{Be}}_{\text{2}}{\text{C+4H}}_{\text{2}}\text{O}\to 2{\text{Be(OH)}}_{\text{2}}{\text{+CH}}_{\text{4}}$

Explanation of Solution

To write balanced equation for ${\text{Be}}_{\text{2}}{\text{C+H}}_{\text{2}}\text{O}\to {\text{Be(OH)}}_{\text{2}}{\text{+CH}}_{\text{4}}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

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

• To balance $\text{Be}$, place two on ${\text{Be(OH)}}_{\text{2}}$ on right hand side

${\text{Be}}_{\text{2}}{\text{C+H}}_{\text{2}}\text{O}\to 2{\text{Be(OH)}}_{\text{2}}{\text{+CH}}_{\text{4}}$

Reactant side	Atom	Product side
2	Be	2
1	C	1
2	H	8
1	O	4

• Place four on water on the left hand side

  ${\text{Be}}_{\text{2}}{\text{C+4H}}_{\text{2}}\text{O}\to 2{\text{Be(OH)}}_{\text{2}}{\text{+CH}}_{\text{4}}$

Reactant side	Atom	Product side
2	Be	2
1	C	1
8	H	8
4	O	4

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{Be}}_{\text{2}}{\text{C+4H}}_{\text{2}}\text{O}\to 2{\text{Be(OH)}}_{\text{2}}{\text{+CH}}_{\text{4}}$

(l)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

Explanation of Solution

To write balanced equation for ${\text{Cu+HNO}}_{\text{3}}\to {\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{+NO+H}}_{\text{2}}\text{O}$

• The charge on copper and on nitrogen is not same on both the sides. This equation can be balanced by redox method.
• Find out and write the redox couples of the reaction. write down it as half reactions.

Oxidation: ${\text{Cu}}^{\text{0}}\to {\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+2e}}^{\text{-}}$

Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+3e^{-}\to {\text{N}}^{\text{(2+)}}\text{O}$

• Balance all the atoms on both sides except oxygen and hydrogen.

Oxidation: ${\text{Cu}}^{\text{0}}+{\text{2HNO}}_{\text{3}}\to {\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+2e}}^{\text{-}}$

Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+3e^{-}\to {\text{N}}^{\text{(2+)}}\text{O}$

• Balance the charge on both the reactions by adding ${\text{H}}^{\text{+}}$

Oxidation: ${\text{Cu}}^{\text{0}}+{\text{2HNO}}_{\text{3}}\to {\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+2e}}^{\text{-}}+{\text{2H}}^{\text{+}}$

Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+3e^{-}+{\text{3H}}^{\text{+}}\to {\text{N}}^{\text{(2+)}}\text{O}$

• Balance the oxygen atoms by adding water molecules.

Oxidation: ${\text{Cu}}^{\text{0}}+{\text{2HNO}}_{\text{3}}\to {\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+2e}}^{\text{-}}+{\text{2H}}^{\text{+}}$

Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+3e^{-}+{\text{3H}}^{\text{+}}\to {\text{N}}^{\text{(2+)}}{\text{O+2H}}_{\text{2}}\text{O}$

• Electron gain must be equivalent to electron lost. This is done by multiplying the coefficients by integers giving lowest common multiple for all the species.

Oxidation: ${\text{3Cu}}^{\text{0}}+6{\text{HNO}}_{\text{3}}\to 3{\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+6e}}^{\text{-}}+6{\text{H}}^{\text{+}}$

Reduction: ${\text{2HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+6e^{-}+6{\text{H}}^{\text{+}}\to 2{\text{N}}^{\text{(2+)}}{\text{O+4H}}_{\text{2}}\text{O}$

• Half reactions are added together and the equation is simplified to get the final balanced equation.

${\text{3Cu}}^{\text{0}}+8{\text{HNO}}_{\text{3}}+{\text{6e}}^{\text{-}}{\text{+6H}}^{\text{+}}\to 3{\text{Cu}}^{\text{(2+)}}{{\text{(NO}}_{\text{3}}\text{)}}_{\text{2}}{\text{+2NO+4H}}_{\text{2}}{\text{O+6e}}^{\text{-}}+6{\text{H}}^{\text{+}}$

• On simplification,

${\text{3Cu+8HNO}}_{\text{3}}\to 3{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{+2NO+4H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
3	C	3
8	H	8
8	N	8
24	O	24

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as ${\text{3Cu+8HNO}}_{\text{3}}\to 3{\text{Cu(NO}}_{\text{3}}{\text{)}}_{\text{2}}{\text{+2NO+4H}}_{\text{2}}\text{O}$

(m)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (m) is $\text{S}+{\text{6HNO}}_{\text{3}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+6NO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{S+HNO}}_{\text{3}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+NO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

• The charge on copper and on nitrogen is not same on both the sides. This equation can be balanced by redox method.
• Find out and write the redox couples of the reaction. write down it as half reactions.

Oxidation: ${\text{S}}^{\text{0}}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6e}}^{\text{-}}$

Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+e^{-}\to {\text{N}}^{\text{(4+)}}{\text{O}}_{\text{2}}$

• Balance all the atoms on both sides except oxygen and hydrogen.

  Oxidation: ${\text{S}}^{\text{0}}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6e}}^{\text{-}}$

  Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+e^{-}\to {\text{N}}^{\text{(4+)}}{\text{O}}_{\text{2}}$

• Balance the charge on both the reactions by adding ${\text{H}}^{\text{+}}$

  Oxidation: ${\text{S}}^{\text{0}}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6e}}^{\text{-}}+{\text{6H}}^{\text{+}}$

  Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+e^{-}+{\text{H}}^{\text{+}}\to {\text{N}}^{\text{(4+)}}{\text{O}}_{\text{2}}$

• Balance the oxygen atoms by adding water molecules.

  Oxidation: ${\text{S}}^{\text{0}}+{\text{4H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6e}}^{\text{-}}+{\text{6H}}^{\text{+}}$

  Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+e^{-}+{\text{H}}^{\text{+}}\to {\text{N}}^{\text{(4+)}}{\text{O}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

• Electron gain must be equivalent to electron lost. This is done by multiplying the coefficients by integers giving lowest common multiple for all the species.

  Oxidation: ${\text{S}}^{\text{0}}+{\text{4H}}_{\text{2}}\text{O}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6e}}^{\text{-}}+{\text{6H}}^{\text{+}}$

  Reduction: ${\text{HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+e^{-}+{\text{H}}^{\text{+}}\to {\text{N}}^{\text{(4+)}}{\text{O}}_{\text{2}}+{\text{H}}_{\text{2}}\text{O}$

• Half reactions are added together and the equation is simplified to get the final balanced equation.

${\text{S}}^{\text{0}}+{\text{6HN}}^{\text{(5+)}}{\text{O}}_{\text{3}}+{\text{4H}}_{\text{2}}{\text{O+6e}}^{\text{-}}{\text{+6H}}^{\text{+}}\to {\text{H}}_{\text{2}}{\text{S}}^{\text{(6+)}}{\text{O}}_{\text{4}}{\text{+6N}}^{\text{(4+)}}{\text{O}}_{\text{2}}{\text{+6H}}_{\text{2}}{\text{O+6e}}^{\text{-}}+{\text{6H}}^{\text{+}}$

• On simplification,

$\text{S}+{\text{6HNO}}_{\text{3}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+6NO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
1	S	1
6	H	6
6	N	6
18	O	18

Now, the atom count on both sides of the reaction are same, hence the equation is balanced. The balanced equation is written as $\text{S}+{\text{6HNO}}_{\text{3}}\to {\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{+6NO}}_{\text{2}}{\text{+2H}}_{\text{2}}\text{O}$

(n)

Interpretation Introduction

Interpretation: The given set chemical equations has to be balanced

Concept Introduction:

• Chemical equation is the symbolic representation of a chemical reaction using the symbols of elements. Chemical equation must be balanced on both sides (reactant side and product side) of the reaction. A balanced equation alone obeys the "Law of conservation of mass"
• Balancing a chemical equation is based on "trial-and-error" method.
• For each step during balancing of chemical equation, the atoms involved in reaction are listed out. Then the number of atoms on both sides of the reaction (reactant side and product side) is checked for tally.
• Suitable numerals are placed as coefficients before the element or compound on either side of reaction wherever necessary.
• Placing of coefficients are allowed for balancing equation, whereas changing the subscripts of elements or compounds are not allowed.
• Always place coefficients for compounds first rather than placing it for elements.
• Polyatomic ions can be treated as single unit on both sides rather than considering their constituents individually.
• While balancing the equation, the numerals used must be simple whole number.
• Balancing the equation can be done by redox method too. This is done by balancing the oxidation number of the atoms involved in equation.

To write: The balanced chemical equation for the given set of reactions

Answer to Problem 3.60QP

The balanced chemical equation for (n) is ${\text{2NH}}_{\text{3}}\text{+3CuO}\to 3{\text{Cu+N}}_{\text{2}}{\text{+3H}}_{\text{2}}\text{O}$

Explanation of Solution

To write balanced equation for ${\text{NH}}_{\text{3}}\text{+CuO}\to {\text{Cu+N}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

• List the atoms of given equation and also the number of each atoms present on both sides of the reaction. Check for the number of atoms available on both sides

Reactant side	Atom	Product side
1	N	2
1	Cu	1
1	O	1
3	H	2

• Multiply ammonia by two on the reactant side

${\text{2NH}}_{\text{3}}\text{+CuO}\to {\text{Cu+N}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	N	2
1	Cu	1
1	O	1
6	H	2

• To balance hydrogen, place three on water

  ${\text{2NH}}_{\text{3}}\text{+CuO}\to {\text{Cu+N}}_{\text{2}}{\text{+3H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	N	2
1	Cu	1
1	O	3
6	H	6

• Finally place three on $\text{CuO}$  and on copper to balance on both the sides.

  ${\text{2NH}}_{\text{3}}\text{+3CuO}\to 3{\text{Cu+N}}_{\text{2}}{\text{+3H}}_{\text{2}}\text{O}$

Reactant side	Atom	Product side
2	N	2
3	Cu	3
3	O	3
6	H	6

Now, the atom count on both sides of the reaction is same, hence the equation is balanced. The balanced equation is written as ${\text{2NH}}_{\text{3}}\text{+3CuO}\to 3{\text{Cu+N}}_{\text{2}}{\text{+3H}}_{\text{2}}\text{O}$