# Balancing of Chemical Equation | Definition, Methods, eg

## Definition of Balanced Chemical Equation

`A balanced chemical equation has equal numbers of atoms for each elements involved in the reaction are represented on the reactant and product sides.`

## Requirements for Balancing The Chemical Equation

The requirements for balancing of chemical equation are:

• This is a requirement that the equation must satisfy to be consistent with the law of conservation of matter.
• It may be confirmed by simply summing the numbers of atoms on either side of the arrow and comparing these sums to ensure they are equal. Note that the number of atoms for a given element is calculated by multiplying the coefficient of any formula containing that element by the element’s subscript in the formula.
• If an element appears in more than one formula on a given side of the equation, the number of atoms represented in each must be computed and then added together.

For example, both product species in the example reaction, CO2 and H2O, contain the element oxygen, and so the number of oxygen atoms on the product side of the equation is

(1CO2 molecule×2O atoms/CO2 molecule )+(2H2O molecule ×1 O atom/H2O molecule )=4 O atoms

• The equation for between methane and oxygen to yield carbon dioxide and water is confirmed to be balanced per this approach, as shown here;

CH4 + 2O2 → CO2 + 2H2O

• Balancing chemical equations involves the addition of stoichiometric coefficients to the reactants and products. This is important because a chemical equation must obey the law of conservation of mass and the law of constant proportions, i.e. the same number of atoms of each element must exist on the reactant side and the product side of the equation.

## The Balancing Methods (How to Balance the Chemical Equation)

The first step that must be followed while balancing chemical equations is to obtain the complete unbalanced equation.

In order to illustrate this method, the combustion reaction between propane and oxygen is taken as an example.

### Step-1

The unbalanced equation must be obtained from the chemical formula of the reactants and the products.

The chemical formula of propane is C3H8. It burns with oxygen (O2) to form carbon dioxide (CO2) and water (H2O).

The unbalanced chemical equation can be written as C3H8 + O2 → CO2 + H2O

### Step-2

The total number of atoms of each element on the reactant side and the product side must be compared.

For example, the number of atoms on each side can be tabulated as follows.

Chemical Equation:

C3H8 + O2 → CO2 + H2O

### Step-3

Now, stoichiometric coefficients are added to molecules containing an element with a different number of atoms in the reactant and product sides. The coefficient must balance the number of atoms on each side.

• Generally, the stoichiometric coefficients are assigned to hydrogen and oxygen atoms last.
• Now, the number of atoms of the elements on the reactant and product side must be updated.
• It is important to note that the number of atoms of an element in one species must be obtained by multiplying the stoichiometric coefficient by the total number of atoms of that element present in 1 molecule of the species.

For example, when coefficient 3 is assigned to the CO2 molecule, the total number of oxygen atoms in CO2 becomes 6. In this example, the coefficient is first assigned to carbon, as tabulated below

Chemical Equation:

C3H8 + O2 → 3CO2 + H2O

### Step-4

Step 3 is repeated until all the number atoms of the reacting elements are equal on the reactant and product side. In this example, hydrogen is balanced next. The chemical equation is transformed as follows.

Chemical Equation:

C3H8 + O2 → 3CO2 + 4H2O

Now that the hydrogen atoms are balanced, the next element to be balanced is oxygen. There are 10 oxygen atoms on the product side, implying that the reactant side must also contain 10 oxygen atoms.

Each O2 molecule contains 2 oxygen atoms. Therefore, the stoichiometric coefficient that must be assigned to the O2 molecule is 5.

The updated chemical equation is tabulated below.

Chemical Equation:

C3H8 + 5O2 → 3CO2 + 4H2O

### Step-5

Once all the individual elements are balanced, each element’s total number of atoms on the reactant and product side are compared once again. If there are no inequalities, the chemical equation is said to be balanced.

In this example, every element now has an equal number of atoms in the reactant and product side. Therefore, the balanced chemical equation is

C3H8 + 5O2 → 3CO2 + 4H2O.

## Examples of Balancing Chemical Equation

Here are some solved examples of balancing chemical equation

### Example-1 of Balancing Chemical Equation

CO2 + H2O → C6H12O6 + O2

The first step is to focus on elements that only appear once on each side of the equation. Here, both carbon and hydrogen fit this requirement.

So, we will start with carbon. There is only one atom of carbon on the left-hand side but six on the right-hand side. So, we add a coefficient of six on the carbon-containing molecule on the left.

6CO2 + H2O → C6H12O6 + O2

Next, let’s look at hydrogen. There are two hydrogen atoms on the left and twelve on the right. So, we will add a coefficient of six on the hydrogen-containing molecule on the left.

6CO2 + 6H2O → C6H12O6 + O2

Now, it is time to check the oxygen. There are a total of 18 oxygen molecules on the left (6×2 + 6×1). On the right, there are eight oxygen molecules.

Now, we have two options to even out the right-hand side: We can either multiply C6H12O6 or O2 by a coefficient. However, if we change C6H12O6, the coefficients for everything else on the left-hand side will also have to change because we will be changing the number of carbon and hydrogen atoms.

To prevent this, it usually helps only to change the molecule containing the fewest elements. In this case, the O2. So, we can add a coefficient of six to the O2 on the right.

6CO2 + 6H2O → C6H12O6 + 6O2

### Example-2

SiCl4 + H2O → H4SiO4 + HCl

The only element that occurs more than once on the same side of the equation here is hydrogen, so that we can start with any other element.

Let’s start by looking at silicon. Notice that there is only one silicon atom on either side, so we do not need to add any coefficients yet.

Next, let’s look at chlorine. There are four chlorine atoms on the left side and only one on the right. So, we will add a coefficient of four on the right.

SiCl4 + H2O → H4SiO4 + 4HCl

Next, let’s look at oxygen. Remember that we first want to analyze all the elements that only occur once on one side of the equation. There is only one oxygen atom on the left but four on the right. So, we will add a coefficient of four on the left-hand side of the equation.

SiCl4 + 4H2O → H4SiO4 + 4HCl

We are almost done! Now, we have to check the number of hydrogen atoms on each side. The left has eight, and the right also has eight, so we are done.

SiCl4 + 4H2O → H4SiO4 + 4HCl

As always, make sure to double-check that the number of atoms of each element balances on each side before continuing.

### Example-3

Al + HCl → AlCl3 + H2

This problem is a bit tricky, so be careful. Whenever a single atom is alone on either side of the equation, it is easiest to start with that element.

So, we will start by counting the aluminum atoms on both sides. One is on the left and one on the right, so we do not need to add any coefficients yet.

Next, let’s look at hydrogen. There is also one on the left but two on the right. So, we will add a coefficient of two on the left.

Al + 2HCl → AlCl3 + H2

Next, we will look at chlorine. There are now two on the left but three on the right. Now, this is not as straightforward as just adding a coefficient to one side. We need the number of chlorine atoms to be equal on both sides. So we need to get two and three to be equal.

We can accomplish this by finding the lowest common multiple. In this case, we can multiply two by three and three by two to get the lowest common multiple of six. So, we will multiply 2HCl by three and AlCl3 by two:

Al + 6HCl → 2AlCl3 + H2

In this case, aluminium is no longer balanced because we added a coefficient to the aluminium-containing molecule on the right-hand side. There is one on the left but two on the right. So, we will add one more coefficient.

2Al + 6HCl → 2AlCl3 + H2

We are not quite done yet. Looking over the equation one final time, we see that hydrogen has also been unbalanced. There are six on the left but two on the right.

2Al + 6HCl → 2AlCl3 + 3H2

### Example-4

Na2CO3 + HCl → NaCl + H2O + CO2

Hopefully, by this point, balancing equations is becoming easier, and you are getting the hang of it. Looking at sodium, we see that it occurs twice on the left but once on the right. So, we can add our first coefficient to the NaCl on the right.

Na2CO3 + HCl → 2NaCl + H2O + CO2

Next, let’s look at carbon. One is on the left and one on the right, so there are no coefficients to add. Since oxygen occurs in more than one place on the left, we will save it for last. Instead, look at hydrogen. One on the left and two on the right, so we will add a coefficient to the left.

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Then, looking at chlorine, we see that it is already balanced with two on each side. Now we can go back to look at oxygen. There are three on the left and three on the right, so our final answer is

Na2CO3 + 2HCl → 2NaCl + H2O + CO2

### Example-5

C7H6O2 + O2 → CO2 + H2O

We can start balancing this equation by looking at carbon or hydrogen. Looking at carbon, we see seven atoms on the left and only one on the right. So, we can add a coefficient of seven on the right.

C7H6O2 + O2 → 7CO2 + H2O

Then, there are six atoms on the left and two on the right for hydrogen. So, we will add a coefficient of three on the right.

C7H6O2 + O2 → 7CO2 + 3H2O

Now, for oxygen, things will get a little tricky. Oxygen occurs in every molecule in the equation.

There are four atoms of oxygen on the left and 17 on the right. There is no obvious way to balance these numbers, so we must use a little trick: fractions.

When writing our final answer, we cannot include fractions as it is not proper form, but it sometimes helps to solve the problem. Also, try to avoid overmanipulating organic molecules.

You can easily identify organic molecules, otherwise known as CHO molecules because they are made up of only carbon, hydrogen, and oxygen. We don’t like to work with these molecules because they are rather complex.

Also, larger molecules tend to be more stable than smaller molecules and less likely to react in large quantities.

So, to balance out the four and seventeen, we can multiply the O2 on the left by 7.5. That will give us

C7H6O2 + 7.5O2 → 7CO2 + 3H2O