Describe the method of balancing a chemical equation taking an example.

Example I: Let us consider the formation of water from the combination of oxygen and hydrogen. The following sequential steps be taken to obtain a balanced chemical equation.

(i) First write the skeleton equation. For example,

H₂+ O₂ → H₂O

Caution: Do not change formula of any constituent while balancing the equation.

(ii)    The same total of charges should appear on the left and right side of the equation.

(iii)    Make one of the atoms equal on both sides by multiplying a molecule or compound with an integral number so that the desired element is balanced. It is convenient to start with the molecule or compound that contains the maximum number of atoms. Here, H₂O contains maximum number of atoms. It contains one short of oxygen atom and so it is multiplied by 2 as shown.    

H_{2 +} O₂ → 2H₂O

(iv)    Next examine the effect of this multiplication of the molecule on the balance of other atoms. As is seen, the oxygen balances but now hydrogen on left is 2 less. So, multiply H₂ on the left by 2

2H₂ + O₂ → 2H₂O

(v)    Further count the number of atoms of each type on both sides. In the above equation

$$\begin{gathered} \mathrm{Right} \mathrm{Left} \\ \mathrm{H} 4 4 \\ \mathrm{O} 2 2 \end{gathered}$$

So, the equation is balanced. If the number of atoms on both sides do not agree, continue with the above steps till Balanced Chemical Equation is obtained. This is called hit and trial method.

Example II: Let us take a little more difficult equation, when iron is combined with steam (H₂O).

(i) The skeleton equation for the above reaction is:
Fe + H₂O → Fe₃O₄ + H₂

(ii) Then, Fe₃O₄ is selected which contains the maximum of atoms. It contains 4 oxygen atoms whereas there is only one oxygen atom on the other side, i.e., L.H.S, in H₂O.
So H₂O is multiplied by 4.
Fe + 4H₂O → Fe₃O₄ + H₂

(iii)    Again examine the effect of step (ii). Oxygen is balanced but Fe and H are not yet balanced. So to balance H, multiply H₂ by 4.

Fe + 4H₂O → Fe₃O₄ + 4H₂

(iv)    When counting the number of atoms on both sides, it is seen Fe is one on L.H.S.
and is 3 on R.H.S. To equalise Fe on both sides, multiply Fe on L.H.S. by 3.

3Fe + 4H₂O → Fe₃O₄ + 4H₂

Now this is a balanced chemical equation.

Example III: Let us consider a reaction in which sodium metal reacts with water to form sodium hydroxide and hydrogen.
(i)    The skeleton equation for the above reaction is:

Na + H₂O → NaOH + H₂

(ii)    When considering number of each type of atoms on both sides, it is seen
hydrogen atoms on both sides are not equal, it is 2 atoms on left side and 3 atoms on right side. So, we multiply H₂O by 2.
Na + 2H₂O → NaOH + H₂

(iii) This step (ii) makes oxygen atoms deficient on right side and hydrogen atoms becomes more on left side. So, we multiply NaOH by 2.
Na + 2H₂O →2NaOH + H₂

(iv) Now oxygen and hydrogen atoms are equal on both sides but sodium atom is one
less on left side. So, we multiply Na by 2.

2Na + 2H₂O → 2NaOH + H₂

Now this is a balanced chemical equation.

Example IV: If the reaction is complicated i.e., it involves large number of reactants and products, it is preferred to write the equation in steps (the actual reaction may or may not be taking in these steps). Each step should be a balanced chemical equation. For example, when copper reacts with conc. nitric acid, products are cupric nitrate, nitrogen dioxide and water.

Cu + HNO₃ Cu(NO₃)₂ + NO₂ + H₂0

We can write this reaction in three steps and balance each step separately.

(i) First, HNO₃ is decomposed to give nitrogen dioxide (NO₂) and water (H₂O) and atomic oxygen.
2HNO₃ → 2NO₂ + H₂O + O    ...(i)

(ii)    In the next step, copper is oxidised to copper (II) oxide (CuO)

Cu + O → CuO    ...(ii)

(iii)    Copper oxide so formed then reacts with nitric acid to form copper nitrate [Cu(NO₃)₂] and water.

CuO + 2HNO₃ → Cu(NO₃)₂ + H₂O    ...(iii)

Next multiply Eq. (i), (ii), (iii) by an integer so that on adding (i), (ii) and (iii), intermediate products (which do not appear in the final reaction) cancel out. Now in this case, we find that the integer is one. So add (i), (ii) and (iii) and write the final balanced equation as shown

Cu + 4HNO₃ → Cu(NO₃)₂ + 2NO₂ + 2H₂O

This method is known as Partial Equation Method.