Chemistry » Physical and Chemical Change » Balancing Chemical Equations

# Balancing Chemical Equations Continued

## Steps to balance a chemical equation through inspection

When balancing a chemical equation, there are a number of steps that need to be followed.

1. Identify the reactants and the products in the reaction and write their chemical formulae.

2. Write the equation by putting the reactants on the left of the arrow and the products on the right.

3. Count the number of atoms of each element in the reactants and the number of atoms of each element in the products.

4. If the equation is not balanced, change the coefficients of the molecules until the number of atoms of each element on either side of the equation balance.

5. Check that the atoms are in fact balanced.

6. (we will look at this a little later): Add any extra details to the equation e.g. phase symbols.

## Example: Balancing Chemical Equations

### Question

Balance the following equation:

$\text{Mg} + \text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}$

### Step 1: Identify the reactants and products

This has been done in the question.

### Step 2: Write the equation for the reaction

This has been done in the question.

### Step 3: Count the number of atoms of each element in the reactants and products

Reactants: $$\text{Mg} = 1 \text{ atom}$$, $$\text{H} = 1 \text{ atom}$$, $$\text{Cl} = 1 \text{ atom}$$

Products: $$\text{Mg} = 1 \text{ atom}$$, $$\text{H} = 2 \text{ atoms}$$, $$\text{Cl} = 2 \text{ atoms}$$

### Step 4: Balance the equation

The equation is not balanced since there are two chlorine atoms in the product and only one in the reactants. If we add a coefficient of two to the $$\text{HCl}$$ to increase the number of $$\text{H}$$ and $$\text{Cl}$$ atoms in the reactants, the equation will look like this:

$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}$

### Step 5: Check that the atoms are balanced

If we count the atoms on each side of the equation, we find the following:

Reactants: $$\text{Mg} = 1 \text{ atom}$$, $$\text{H} = 2 \text{ atoms}$$, $$\text{Cl} = 2 \text{ atoms}$$

Products: $$\text{Mg} = 1 \text{ atom}$$, $$\text{H} = 2 \text{ atoms}$$, $$\text{Cl} = 2 \text{ atoms}$$

The equation is balanced. The final equation is:

$\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_{2} + \text{H}_{2}$

## Example: Balancing Chemical Equations

### Question

Balance the following equation:

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

### Step 1: Count the number of atoms of each element in the reactants and products

Reactants: $$\text{C} = 1$$, $$\text{H} = 4$$, $$\text{O} = 2$$

Products: $$\text{C} = 1$$, $$\text{H} = 2$$, $$\text{O} = 3$$

### Step 2: Balance the equation

If we add a coefficient of $$\text{2}$$ to $$\text{H}_{2}\text{O}$$, then the number of hydrogen atoms in the products will be $$\text{4}$$, which is the same as for the reactants. The equation will be:

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

### Step 3: Check that the atoms balance

Reactants: $$\text{C} = 1$$, $$\text{H} = 4$$, $$\text{O} = 2$$

Products: $$\text{C} = 1$$, $$\text{H} = 4$$, $$\text{O} = 4$$

You will see that, although the number of hydrogen atoms now balances, there are more oxygen atoms in the products. You now need to repeat the previous step. If we put a coefficient of $$\text{2}$$ in front of $$\text{O}_{2}$$, then we will increase the number of oxygen atoms in the reactants by $$\text{2}$$. The new equation is:

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

When we check the number of atoms again, we find that the number of atoms of each element in the reactants is the same as the number in the products. The equation is now balanced.

## Example: Balancing Chemical Equations

### Question

In our bodies, sugar $$(\text{C}_{6}\text{H}_{12}\text{O}_{6})$$ reacts with the oxygen we breathe in to produce carbon dioxide, water and energy. Write the balanced equation for this reaction.

### Step 1: Identify the reactants and products in the reaction.

Reactants: sugar $$(\text{C}_{6}\text{H}_{12}\text{O}_{6})$$ and oxygen $$(\text{O}_{2})$$

Products: carbon dioxide $$(\text{CO}_{2})$$ and water $$(\text{H}_{2}\text{O})$$

### Step 2: Write the equation

$\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow \text{CO}_{2} + \text{H}_{2}\text{O}$

### Step 3: Count the number of atoms of each element in the reactants and in the products

Reactants: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 8$$

Products: $$\text{C} = 1$$, $$\text{H} = 2$$, $$\text{O} = 3$$

### Step 4: Balance the equation

It is easier to start with carbon as it only appears once on each side. If we add a $$\text{6}$$ in front of $$\text{CO}_{2}$$, the equation looks like this:

$\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow 6\text{CO}_{2} + \text{H}_{2}\text{O}$

Reactants: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 8$$

Products: $$\text{C} = 6$$, $$\text{H} = 2$$, $$\text{O} = 13$$

### Step 5: Change the coefficients again to try to balance the equation.

Let us try to get the number of hydrogens the same this time.

$\text{C}_{6}\text{H}_{12}\text{O}_{6} + \text{O}_{2} \rightarrow 6\text{CO}_{2} + 6\text{H}_{2}\text{O}$

Reactants: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 8$$

Products: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 18$$

### Step 6: Now we just need to balance the oxygen atoms.

$\text{C}_{6}\text{H}_{12}\text{O}_{6} + 6\text{O}_{2} \rightarrow 6\text{CO}_{2} + 6\text{H}_{2}\text{O}$

Reactants: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 18$$

Products: $$\text{C} = 6$$, $$\text{H} = 12$$, $$\text{O} = 18$$