How do you balance a chemical equation?

Balanced Equations

When an equation is balanced, there are the same numbers of each type of atom on the reactant and product sides of the equation.

For example, the following equation is balanced because there are four Al atoms and six O atoms on both sides of the equation.

`~4Al + 3~o_2 -gt ~2Al_2O_3 `

 Reactants              Products

4 Al atoms             4 Al atoms

6 O atoms              6 O atoms

How to Balance an Equation

It is important to remember what you can and can’t do when balancing equations.

• You can add coefficients in front of substances.


• You cannot add coefficients in the middle of substances.


• You cannot change subscripts.

Let’s look at an example of how to balance an equation. The following equation is unbalanced.

`~Au_2S_3 + ~H_2 -gt Au + ~H_2S `

Step 1: Make a list of how many atoms of each element are present on the reactant and product sides of the equation.

`~Au_2S_3 + ~H_2 -gt Au + ~H_2S`

Reactants              Products

Au: 2                    Au 1

S: 3                      S: 1

H: 2                      H: 2

Step 2: Choose one of the unbalanced atoms to start with.

Let’s start with Au. There are two Au atoms on the reactant side, but only one Au atom on the product side.

Remember, when a subscript is not indicated for an atom, the subscript is actually “1”. i.e.  Au is the same as `~Au_1` .

Ask yourself “What number can I multiply with the subscript of the Au atom, on the product side of the equation, that will give me the same number of Au atoms as are on the reactant side of the equation?” The answer would be 2.

So, we will add a coefficient of 2 in front of the Au atom on the product side of the equation.

`~Au_2S_3 + ~H_2 -gt 2Au + ~H_2S`

Reactants              Products

Au: 2                      Au 2

S: 3                        S: 1

H: 2                        H: 2

Notice that there are now two Au atoms on both sides of the equation.

Step 3: Choose another unbalanced atom.

The only other currently unbalanced atom is S.

Ask yourself “What number can I multiply with the subscript of the S on the product side of the equation, that will give me the same number of S atoms as are on the reactant side of the equation?” The answer would be 3.

So, we will add a coefficient of 3 in front of the compound containing the S atom on the product side of the equation.

`~Au_2S_3 + ~H_2 -gt 2Au + ~3H_2S`

Reactants              Products

Au: 2                      Au 2

S: 3                        S: 3

H: 2                        H: 6

Notice that the coefficient is placed in front of the entire `~H_2S` compound. It is NOT placed in the middle of the compound. This changes the number of S atoms on the product side to three AND it changes the number of H atoms on the product side to 6. So, now the H atoms are unbalanced!

Step 4: Balance the last unbalanced atom.

The last unbalanced atom is H.

Ask yourself “What number can I multiply with the subscript of the H on the reactant side of the equation that will give me the same number of H atoms as are on the product side of the equation?" The answer would be “3”.

So, we will add a coefficient of 3 in front of the H atoms on the reactant side of the equation.

`~Au_2S_3 + ~3H_2 -gt 2Au + ~3H_2S`

Reactants              Products

Au: 2                      Au 2

S: 3                        S: 3

H: 6                        H: 6

The equation is now balanced.