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Chemistry > Chemical Reactions and Equations > How to Balance Chemical Equations? | 5 Simple and Easy Steps
Chemical Reactions and Equations

How to Balance Chemical Equations? | 5 Simple and Easy Steps

Symbolic representation of a chemical reaction is referred to as a chemical equation. Basically, the symbols are the letter that represents an element. Moreover, in a chemical reaction, the reactants are written on the left side and products are written on the right side. In addition, we connect these two sides with an arrow leading from the left to the right, symbolizing the reaction. Besides, in this article, we will learn how to balance chemical equations?

As per the law of conservation of mass, “atom can neither be created nor can be destroyed”. So, the product and reactants of chemical equations should have an equal number of atoms on both sides.

Methods of Balancing Chemical Equations

  1. Traditional Balance Method
  2. Completing Algebraic Balance Method

how to balance chemical equations

Method 1 – Traditional Balance

  1. Write down the given equation

\(C_{8}H_{8} + O_{2} \rightarrow H_{2}O + CO_{2}\)

In this reaction, propane burns in the presence of oxygen to produce water and carbon dioxide.

  1. Write the number of atoms per element

Look at the subscripts numbers written next to each element to find the number of atoms in the equation. Make sure to do this for both sides. Also, it would be better to connect it back to the original equation, noting how each element appears.

For instance, in the above example, you have 3 oxygen on the right side, but that total results from addition. The left side has 3 carbon, 8 hydrogens, and 2 oxygen. While, the right side has 1 carbon, 2 hydrogens, and 3 oxygen.

  1. Save oxygen and hydrogen for the last

Both oxygen and hydrogen are common in molecules. So, it’s likely that you’ll have them on both sides of your equation because it’s better to balance them last. Also, you need to recount your atoms for balancing the oxygen and hydrogen as you’ll likely need to use a coefficient to balance the other atoms.

  1. Begin with a single element

Firstly, if you have more than one element to balance then select the one that appears in a single molecule of reactants and in a single molecule of products. In this, you need to balance the carbon atom first.

  1. Use coefficient to balance carbon atom

For this add a coefficient before the single carbon atom on the right side of the equation to balance it off with 3 carbon atoms on the left of the equation.

\(C_{3}H_{8} + O{2} \rightarrow H_{2}O + 3 CO_{2}\)

Now you a see that on both sides there are 3 carbon atoms which make carbon atom in balance. However, in a chemical equation, you can alter the coefficient but you can’t change the subscripts.

  1. Next, balance the hydrogen

As you have to balance all the atoms so next balance the hydrogen. On the left side, there is 8 hydrogen and in the right, there are only 2. So, use coefficient to balance hydrogen by adding coefficient 4 before the hydrogen atom as it already has subscript 2. Hence there are 8 hydrogen atoms on both the side as when you multiply confident 4 with subscript 2 then you get 8.

\(C_{3}H_{8} + O_{2} \rightarrow 4H_{2}O + 3CO_{2}\)

  1. Balancing oxygen

According to the general rule, coefficient multiplies with a subscript. In this way on the right side there are 10 oxygen atoms (\(3CO_{2}\rightarrow 3×2= 6 + 4 atoms from 4H_{2}O\)). So to balance it, you have to add 8 more atoms on the left side.

For doing this, add a coefficient 5 to the oxygen molecule to make it 10 (5×2= 10).

\(C_{3}H_{8} +5O_{2} \rightarrow 4H_{2}O + 3CO_{2}\)

Now, on both the side there is equal number of carbon, hydrogen, and oxygen atoms. Hence, the equation is complete.

Method 2- Completing Algebraic Balance

Algebraic balance method or Bottomley’s method, is very useful for complex reactions however, it takes a bit longer.

  1. Write down the equation

Firstly, write down the equation as it is

\(PCl_{5} + H_{2}O \rightarrow H_{3}PO_{4} + HCl\)

  1. Allot letter to each substance

Now allot a letter to each compound or element

\(aPCl_{5} + bH_{2}O \rightarrow cH_{3}PO_{4} + dHCl\)

  1. Check the numbers

See if the elements exist on both sides and set those equal to each other.

\(aPCl_{5} + bH_{2}O \rightarrow cH_{3}PO_{4} + dHCl\)

In the equation, on the left side there are 2b atoms of hydrogen (2 for every molecule of \(H_{2}O\)), While on the right side, there are 3c+d atoms of hydrogen (3 for every molecule of \(H_{3}PO_{4}\) and 1 for every molecule of HCl). Also, the number of atoms of hydrogen has to be equal on both sides, 2b need to be equal to 3c+d.

  1. Now repeat this with each element.

P: a = c

Cl: 5a = d

H: 2b = 3c + d

  1. Solve the numeric equation

Now solve all these numeric equations to get values of each coefficient. As there are more variables than equations so there can be multiple solutions. Al you need to do is find the smallest and non-fractional form. For doing it quickly assign a value for the variable. Let’s take a=1. Then,

As P: a = c, we know that c = 1.

Subsequently Cl: 5a = d, we know that d = 5

Since H: 2b = 3c + d, now you can calculate b like this:

2b = 3(1) + 5

2b = 3 + 5

2b = 8

b=4

So this displays that the values are as:

a = 1

b = 4

c = 1

d = 5

So the complete equation is:

\(PCl_{5} + 4H_{2}O \rightarrow H_{3}PO_{4} + 5HCl\)

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