### Theory:

Can your recall the concept called 'law of conservation of mass' from class IX? Mass cannot be created nor destroyed in a chemical reaction. The entire mass of the elements present in the products of a chemical reaction must be similar to the total mass of the elements present in the reactants.
In other words, before and after a chemical reaction, the number of atoms in each element remains the same. As a result, a skeletal chemical equation must be balanced.

The word-equation for change in temperature and evolution of hydrogen may be expressed by

Zinc + Sulphuric acid → Zinc sulphate + Hydrogen

The following chemical equation may describe the above word-equation -

$$Zn$$ + $$H_2SO_4$$ → $$ZnSO_4$$ + $$H_2$$

Let us analyse the number of atoms of various elements on both sides of the arrow.

 Element Number of atoms in reactants (LHS) Number of atoms in products (RHS) $$Zn$$ $$1$$ $$1$$ $$H$$ $$2$$ $$2$$ $$S$$ $$1$$ $$1$$ $$O$$ $$4$$ $$4$$

Because the number of atoms in each element is the same on both sides of the arrow, the above equation is a balanced chemical equation.
Example:
Let us try to balance the following chemical equation.

$$Fe$$ + $$H_2O$$ → $$Fe_3O_4$$ + $$H_2$$

Step 1: First, draw boxes around each formula to balance a chemical equation. While balancing the equation, do not make any changes inside the boxes.

Step 2: Write the number of atoms of each element in the unbalanced (above) equation.

 Element Number of atoms in reactants (LHS) Number of atoms in products (RHS) $$Fe$$ $$1$$ $$3$$ $$H$$ $$2$$ $$2$$ $$O$$ $$1$$ $$4$$

Step 3: It is typically easier to start balancing with the compound that includes the maximum number of atoms. It could be a reactant or product. Choose the element which has the maximum number of atoms in that compound. Based on these factors, we chose $$Fe_3O_4$$ and the element oxygen. On the RHS, there are four oxygen atoms, whereas, on the LHS, there is just one.

To balance the oxygen atoms-

 Atoms of Oxygen In reactants In products (i) Initial $$1$$ (in $$H_2O$$) $$4$$ (in $$Fe_3O_4$$) (ii) To balance $$1$$ × $$4$$ $$4$$

To balance the number of atoms, we must remember that we cannot change the formulae of the compounds or elements included in the reactions. For instance, to balance oxygen atoms, we can insert coefficient ‘$$4$$’ as $$4H_2O$$ and not $$H_2O_4$$ or ($$H_2O)_4$$.
Now the partly balanced equation becomes -

(partly balanced equation)

Step 4: The atoms of $$Fe$$ and $$H$$ are still not balanced. Choose any of these elements to move further. Let us first balance hydrogen atoms in the partially balanced equation.
To balance the number of $$H$$ atoms, make the number of hydrogen molecules four on the RHS.

 Atoms of hydrogen In reactants In products (i) Initial $$8$$ (in $$4$$ $$H_2O$$) $$2$$ (in $$H_2$$) (ii) To balance $$8$$ $$2$$ × $$4$$

The equation would be -
(partly balanced equation)

Step 5: Examine the above equation for the unbalanced third element. You discover that only one element is left to be balanced, that is, iron $$Fe$$.

 Atoms of iron In reactants In products (i) Initial $$1$$ (in $$Fe$$) $$3$$ (in $$Fe_3O_4$$) (ii) To balance $$1$$ × $$3$$ $$3$$

To equalise $$Fe$$, we place three $$Fe$$ atoms on the LHS.

Step 6: Finally, on both sides of the equation, we count the atoms of each element to confirm that the balanced equation is true.

$$3Fe$$ + $$4H_2O$$ → $$Fe_3O_4$$ + $$4H_2$$               (balanced equation)

Each side of the above equation has the same number of atoms of each element. This equation is now balanced. This method of balancing chemical equations is known as the 'hit-and-trial' method because we conduct trials to balance the equation with the smallest whole number coefficient.
Step 7: Writing symbols of physical states

Examine the above mentioned balanced equation carefully. Is there anything we can learn from this equation about the physical states of each reactant and product? This equation contains no information about their physical states.
The physical states of the reactants and products and their chemical formulae are given in a chemical equation to make it more informative. The aqueous, liquid, gaseous and solid states of reactants and products are expressed by the notations ($$aq$$), ($$l$$), ($$g$$) and ($$s$$), respectively. The word aqueous ($$aq$$) is written if the reactant or product exists as a solution in water.

$$3Fe$$(s) + $$4H_2O$$(g) → $$Fe_3O_4$$(s) + $$4H_2$$(g)

The symbol ($$g$$) with $$H_2O$$ indicates that water is used in the form of steam in this reaction. A chemical equation generally does not incorporate physical states unless they are necessary to be described.

The reaction conditions, such as pressure, catalyst, temperature, etc., are sometimes indicated above and below the arrow in the equation.

For example -
${\mathit{CO}}_{\left(g\right)}\phantom{\rule{0.147em}{0ex}}+{2H}_{2\left(g\right)}\phantom{\rule{0.147em}{0ex}}\stackrel{340\phantom{\rule{0.147em}{0ex}}\mathit{atm}}{⟶}\phantom{\rule{0.147em}{0ex}}{\mathit{CH}}_{3}{\mathit{OH}}_{\left(l\right)}$

(Glucose)