Balancing Chemical Equations Systematically

## **Problem**

Consider the following chemical reactions:

(1)

CH

4

O

2

⟶

CO

2

H

2

O

{\displaystyle \ce{CH4 + O2 -> CO2 + H2O}}

(2)

MoS

2

O

2

⟶

MoO

3

SO

2

{\displaystyle \ce{MoS2 + O2 -> MoO3 + SO2}}

(3)

HClO

4

P

4

O

10

⟶

H

3

PO

4

Cl

2

O

7

{\displaystyle \ce{HClO4 + P4O10 -> H3PO4 + Cl2O7 }}

Name each chemical compound and balance each chemical reaction.

## **Solution**

(1)

CH

4

O

2

⟶

CO

2

H

2

O

{\displaystyle \ce{CH4 + O2 -> CO2 + H2O}}

This is a combustion reaction with methane and diatomic oxygen gas producing carbon dioxide and water.

Introduce variables for the coefficients:

aCH

4

bO

2

⟶

cCO

2

dH

2

O

{\displaystyle \ce{aCH4 + bO2 -> cCO2 + dH2O}}

Derive a system of linear equations for each element

**Carbon:**

# a
c

{\displaystyle a=c}

**Hydrogen:**

4
# a
2
d

{\displaystyle 4a = 2d}

**Oxygen:**

2
# b
2
c
d

{\displaystyle 2b = 2c + d}

It can be deduced that

# a
c

{\displaystyle a=c}

 and

# b
d

{\displaystyle b=d}

 where

# a
c
<
# b
d

{\displaystyle a = c < b = d}

, so set the smallest variables equal to 1. Subsequently,

# c
1

[
5
p
t
]
4
⋅
# 1
2
d
⇒
# d
2

2
# b
2
⋅
1
2
⇒
# b
2

{\displaystyle \begin{align}
a = c = 1 \\ [5 pt]
4\cdot 1 = 2d \Rightarrow d = 2 \\[5 pt]
2b = 2\cdot 1 + 2 \Rightarrow b = 2
\end{align}}

The balanced chemical equation is

CH

4

2

O

2

⟶

CO

2

2

H

2

O

{\displaystyle \ce{CH4 + 2O2 -> CO2 + 2H2O}}

(2)

MoS

2

O

2

⟶

MoO

3

SO

2

{\displaystyle \ce{MoS2 + O2 -> MoO3 + SO2}}

This reaction involves reacting molybdenum disulfide with diatomic oxygen gas to produce molybdenum trioxide with sulfer dioxide .

Introduce variables for the coefficients:

aMoS

2

bO

2

⟶

cMoO

3

dSO

2

{\displaystyle \ce{aMoS2 + bO2 -> cMoO3 + dSO2}}

Derive a system of linear equations for each element

**Molybdenum:**

# a
c

{\displaystyle a=c}

**Sulfer:**

2
# a
d

{\displaystyle 2a = d}

**Oxygen:**

2
# b
3
c
2
d

{\displaystyle 2b = 3c + 2d}

It can be deduced that

# a
c
<
d
<
b

{\displaystyle a = c < d < b}

, so set the smallest variables equal to 1. Subsequently,

# c
1

[
5
p
t
]
2
⋅
# 1
d
⇒
# d
2

2
# b
3
⋅
1
2
⋅
2
⇒
# b

7
2

{\displaystyle \begin{align}
a = c = 1 \\ [5 pt]
2\cdot 1 = d \Rightarrow d = 2 \\[5 pt]
2b = 3\cdot 1 + 2 \cdot 2 \Rightarrow b = \frac{7}{2}
\end{align}}

Substituting the coefficients back into the chemical equation gives

aMoS

2

bO

2

⟶

cMoO

3

dSO

2

{\displaystyle \ce{aMoS2 + bO2 -> cMoO3 + dSO2}}

We cannot have fractional coefficients; thus, the entire equation need to be scaled by 2. The balanced chemical equation is

2

MoS

2

7

O

2

⟶
2

MoO

3

4

SO

2

{\displaystyle \ce{2MoS2 + 7O2 -> 2MoO3 + 4SO2}}

(3)

HClO

4

P

4

O

10

⟶

H

3

PO

4

Cl

2

O

7

{\displaystyle \ce{HClO4 + P4O10 -> H3PO4 + Cl2O7 }}

This reaction involves reacting perchloric acid with phosphorus pentoxide to produce phosphoric acid with dichlorine heptoxide.

Introduce variables for the coefficients:

aHClO

4

bP

4

O

10

⟶

cH

3

PO

4

dCl

2

O

7

{\displaystyle \ce{aHClO4 + bP4O10 -> cH3PO4 + dCl2O7 }}

Derive a system of linear equations for each element

**Hydrogen:**

# a
3
c

{\displaystyle a = 3c}

**Chlorine:**

# a
2
d

{\displaystyle a = 2d}

**Oxygen:**

4
a
10
# b
4
c
7
d

{\displaystyle 4a + 10b = 4c + 7d}

**Phosphorus:

4
# b
c

{\displaystyle 4b = c}**

It can be deduced that

b
<
c
<
d
<
a

{\displaystyle b < c < d < a}

, so set the smallest variables equal to 1. Subsequently,

# b
1

[
5
p
t
]
4
⋅
# 1
c
⇒

# a
3
⋅
4
⇒

# 12
2
d
⇒
# d
6

{\displaystyle \begin{align}
b = 1 \\ [5 pt]
4\cdot 1 = c \Rightarrow c = 4 \\[5 pt]
a = 3\cdot 4 \Rightarrow a = 12 \\[5 pt]
12 = 2d \Rightarrow d = 6
\end{align}}

The balanced chemical equation is

12

HClO

4

P

4

O

10

⟶
4

H

3

PO

4

6

Cl

2

O

7

{\displaystyle \ce{12HClO4 + P4O10 -> 4H3PO4 + 6Cl2O7 }}