5.2: Balancing Chemical Equations

As noted earlier, a balanced chemical equation shows the same number of each kind of atom on both sides of the equation. The process of balancing chemical equations is an important exercise in chemistry and is addressed here.

Consider a simple example of balancing a chemical equation, the reaction of methane, CH₄, with elemental chlorine, Cl₂, to produce dichloromethane, CH₂Cl₂, an important laboratory solvent, and byproduct hydrogen chloride, HCl. The unbalanced chemical equation is

\[\ce{CH4 + Cl2 \rightarrow CH2Cl2 + HCl}\]

Inspection of this equation as it is written shows that it is not balanced because it has 4 H on the left, but just 3 on the right and 2 Cl on the left, but 3 Cl on the right. In order to balance such an equation, consider one element at a time. Carbon is already balanced, so it is best to avoid changing any of the numbers in front of the C-containing compounds. The equation can be balanced for H by putting a 2 in front of HCl:

\[\ce{CH4 + Cl2 \rightarrow CH2Cl2 + 2HCl}\]

Now everything is balanced except for Cl, of which there are 4 on the right, but just 2 on the left.Placing a 2 in front of Cl, gives the required 4 Cls on the left:

\[\ce{CH4 + 2Cl2 \rightarrow CH2Cl2 + 2HCl}\]

This equation is now balanced with 1 C, 4 Hs, and 4 Cls on both the left and the right.

A crucial thing to remember in balancing a chemical equation is that the chemical formulas must not be altered. Only the relative numbers of reactant and product species may be changed.

Next consider the reaction of methane, CH₄, with iron oxide, Fe₂O₃, to give iron metal, Fe, carbon dioxide, CO₂, and water, H₂O. The unbalanced equation is

\[\ce{CH4 + Fe2O3 \rightarrow Fe + CO2 + H2O}\]

In this case it is helpful to note that CH₄ is the only source of both C and H and that 4 times as many H atoms as C atoms must appear in the products. That means that for each CO₂ there must be 2 H₂Os. Both C and H are balanced in the following:

\[\ce{CH4 + Fe2O3 \rightarrow Fe + CO2 + 2H2O}\]

But now O is not balanced. Furthermore, the 3 Os in Fe₂O₃ means that the number of O atoms must be divisible by 3, so try multiplying the three species balanced so far — CH₄, CO₂, and 2H₂O — by 3:

\[\ce{3CH4 + Fe2O3 \rightarrow Fe + 3CO2 + 6H2O}\]

That gives a total of 12 O atoms on the right, 6 each in 3 CO₂ and 6 H₂O. Taking 4 times Fe₂O₃ gives 12 Os on the left:

\[\ce{3CH4 + 4Fe2O3 \rightarrow Fe + 3CO2 + 6H2O}\]

The only species remaining to be balanced is Fe, which can be balanced by putting 8 in front of Fe on the right. The balanced equation is

\[\ce{3CH4 + 4Fe2O3 \rightarrow 8Fe + 3CO2 + 6H2O}\]

Checking the answer shows on both left and right 3 C, 8 Fe, 12 H, and 12 O demonstrating that the equation is in fact balanced.

Exercise: Balance the following:

1. \(\ce{Fe2O3 + CO \rightarrow Fe + CO2}\)

2. \(\ce{FeSO4 + O2 + H2O \rightarrow Fe(OH)3 + H2SO4}\)

3. \(\ce{C2H2 + O2 \rightarrow CO2 + H2O}\)

4. \(\ce{Mg3N2 + M2O \rightarrow Mg(OH)2 + NH3}\)

5. \(\ce{NaAlH4 + H2O \rightarrow H2 + NaOH + Al(OH)3}\)

6. \(\ce{Zn(C2H5)2 + O2 \rightarrow ZnO + CO2 + H2O}\)

Answers: (1) \(\ce{Fe2O3 + 3CO \rightarrow 2Fe + 3CO2}\), (2) \(\ce{4FeSO4 + O2 + 10H2O \rightarrow 4Fe(OH)3 + 4H2SO4}\), (3) \(\ce{2C2H2 + 5O2 \rightarrow 4CO2 + 2H2O}\), (4) \(\ce{Mg3N2 + 6H2O \rightarrow 3Mg(OH)2 + 2NH3}\) (5) \(\ce{NaAlH4 + 4H2O \rightarrow 4H2 + NaOH + Al(OH)3}\), (6) \(\ce{Zn(C2H5)2 + 7O2 \rightarrow ZnO + 4CO2 + 5H2O}\)