A chemical equation is the representation of the chemical reaction taking place. This gives us the exact moles of reactants reacting and the products formed. So,** it is important to provide the exact moles of the reacting substances and products.** Also, we know that according to the law of mass action, moles of a product formed would be equal to the moles of the reactant. Nothing will be lost in the process of a chemical reaction.
Keeping this into account, we need to balance a chemical equation for providing the exact picture of a chemical reaction.

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The number of atoms of each element is conserved during a chemical reaction. A correctly written chemical equation reflects this observation. To satisfy the law of conservation of mass, we need to insert the right coefficients into a chemical equation.

**A general method to learn the balancing of chemical equations is as follows :**

**Example** :

Balance the chemical equation:

**Fe + H**_{2}O $\rightarrow $** Fe**_{3}O_{4} + H_{2}

We can see that the equation is not already balanced. **3 Fe + H**_{2}O $\rightarrow $ **1 Fe**_{3}O_{4} + H_{2}

**3**** Fe + 4H**_{2}O $\rightarrow $ **1 Fe**_{3}O_{4} + H_{2}

**3 Fe + 4H**_{2}O $\rightarrow $ **1 Fe**_{3}O_{4} + 4H_{2}

Thus, the following equation is completely balanced, for all the atoms.

### Solved Examples

**Question 1: **Balance the following chemical equation: NH3 + O2 → NO + H2O

** Solution: **

Let us write the unbalanced equation first:

NH_{3} + O_{2 }→ NO + H_{2}O

Now, counting the number of each atoms on both sides

Except for hydrogen, the other two elements present are balanced. Let us start with balancing the hydrogen

Since hydrogen occurs in only one substance, in both products and reactant side, and also, since in one side, its number is an odd number, 3 and on the other side, there are 2 numbers, which is even, let us take a common multiple and make the number of hydrogen on both sides as '6'. And also, satisfying the nitrogen, by adding a 2 before NO:

2NH_{3 }+ O_{2} → 2 NO + 3 H_{2}O

On doing this, the number of oxygen and nitrogen changes too. And we have the same odd and even problem now with oxygen too. There are 2 oxygen on the reactants side and 5 on the product side.

By doubling all coefficients to make it even:

[2NH_{3} + O_{2} → 2 NO + 3 H2O ] x 2

4NH_{3} + 2O2 → 4 NO + 6 H2O

Now, finally, we see that oxygen is still unbalanced. It is less on the reactant side. Changing the coefficient of oxygen on the left-hand side, the reactants side, we get

4NH_{3} + 5O_{2} → 4 NO + 6 H_{2}O

The equation is finally balanced.

**Question 2: **Write the balanced chemical reaction for the combustion of glucose, C_{6}H_{12}O_{6} In air.

** Solution: **

This process is referred to as balancing an equation. Trial and error is the best way to learn how to balance equations. There are an equal number of atoms for each element on each side of the equation, when the equation is balanced.

- Write down the unbalanced equation containing the correct molecular formula of all reactants and products.
- Determine whether or not the chemical equation is already balanced.
- If the equation is not balanced, start by balancing the element that occurs in the fewest number of reactants and the product molecules.
**Start with carbon or any other atom with a high molar mass.** - Balance the remaining elements.
- Make sure that the number of atoms in each element is balanced.

Balance the chemical equation:

We can see that the equation is not already balanced.

- We start by balancing Fe, since it has a high molecular mass atom and only appears once on the reactant and the product side. We see that there are three Fe atoms on the product side, so, we can balance Fe by adding a coefficient '3' to Fe on the reactant side. So:

- Having balanced Fe, we can then turn to oxygen, since it occurs with Fe in Fe
_{3}O_{4}, and we have already balanced Fe. There are four atoms of oxygen on the product side, so we can balance oxygen by placing a coefficient of 4 in front of H2O on the reactants side.

- Finally, Hydrogen is the only one, left unbalanced. We have 8 hydrogen atoms on the left side, in water and so, we need 8 on the right side too. By placing '4' in front of H
_{2}, we will get 8 hydrogen.

Thus, the following equation is completely balanced, for all the atoms.

Though, there are many types of chemical reactions,

all reactions are balanced in the same way, except for redox reactions, which require a different method of balancing.

Let us start with a chemical equation for a reaction involving the combustion of hydrogen gas with oxygen gas. The two gases are ignited and the products formed are water and energy.

Write the formula for balancing the chemical equations for the reactants and products.

**H _{2 }+ O_{2 }**→

The number of atoms in a chemical equation can be equally balanced. Two oxygen atoms can be presented in the reactant side but on the product side only one atom is presented here.

To solve the equations, on the product side, place 2 before the formula for water.

** H _{2} + O_{2} **→

Now, we can see, two molecules of water can be produced for each molecule of oxygen that can be reacted. The number is placed before the formula and is called the chemical equation solver’s coefficient.

A coefficient represents the number to be multiplied by everything in the formula for both the reactant and the product sides. The water molecule on the product side is doubled so that the two oxygen atoms and four hydrogen atoms combined, on the reactant side, equal the number of atoms on the product side.

For balancing the equation, on the reactant side, place 2 in front of the H_{2}.

**2H _{2} + O_{2 }**→

Hence, there are four hydrogen atoms and 2 oxygen atoms represented on both the reactant and the product side.

Atoms of Reactants |
Atoms of products |

4H | 4H |

2O | 2O |

For the chemical equation balancing solver for water molecule, the atoms involved can neither be created nor destroyed.

Let us write the unbalanced equation first:

NH

Now, counting the number of each atoms on both sides

Elements |
Reactants side |
Product side |

Nitrogen | 1 | 1 |

Oxygen | 2 | 2 |

Hydrogen | 3 | 2 |

Except for hydrogen, the other two elements present are balanced. Let us start with balancing the hydrogen

Since hydrogen occurs in only one substance, in both products and reactant side, and also, since in one side, its number is an odd number, 3 and on the other side, there are 2 numbers, which is even, let us take a common multiple and make the number of hydrogen on both sides as '6'. And also, satisfying the nitrogen, by adding a 2 before NO:

2NH

On doing this, the number of oxygen and nitrogen changes too. And we have the same odd and even problem now with oxygen too. There are 2 oxygen on the reactants side and 5 on the product side.

By doubling all coefficients to make it even:

[2NH

4NH

Now, finally, we see that oxygen is still unbalanced. It is less on the reactant side. Changing the coefficient of oxygen on the left-hand side, the reactants side, we get

4NH

The equation is finally balanced.

To solve this question, we start by showing the unbalanced reaction

**C**_{6}H_{12}O_{6} + O_{2} →** CO**_{2} + H_{2}O

We can start by balancing carbon. There are 6 C atoms on the left-hand side, so we can place a coefficient of 6 in front of CO_{2}, on the right-hand side. This gives us:

**C**_{6}H_{12}O_{6} + O_{2} →** 6 CO**_{2} + H_{2}O

Now, we can move to hydrogen. There are 12 H atoms on the left so we can add a coefficient of 6 in front of water.

**C**_{6}H_{12}O_{6} + O_{2} →** 6 CO**_{2} + 6H_{2}O

Now, only oxygen is left to be balanced. We have a total of 12 + 6 or 18 oxygen on the right-hand side.

In the left-hand side, we have 6 oxygen in glucose molecule.

Therefore, we need to make sure that there are 12 more oxygen atoms on the left, which we can accomplish by adding a coefficient of 6 in front of Oxygen.

**C**_{6}H_{12}O_{6} +6 O_{2} →** 6 CO**_{2} + 6H_{2}O

Thus, the equation is completely balanced.

We can start by balancing carbon. There are 6 C atoms on the left-hand side, so we can place a coefficient of 6 in front of CO

Now, we can move to hydrogen. There are 12 H atoms on the left so we can add a coefficient of 6 in front of water.

Now, only oxygen is left to be balanced. We have a total of 12 + 6 or 18 oxygen on the right-hand side.

In the left-hand side, we have 6 oxygen in glucose molecule.

Therefore, we need to make sure that there are 12 more oxygen atoms on the left, which we can accomplish by adding a coefficient of 6 in front of Oxygen.

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