Question

How to balance the equation by shortcut method?

Answer 1

Hint : As the law of conservation of mass states that “mass can neither be created nor be destroyed” so during a reaction when we write an equation then some of the product on the reactant side may be more or less than the product formed. To follow the law of mass and also constant proportion we need to balance a chemical reaction.

Complete Step By Step Answer:
To understand the balancing of equations more clearly let’s take an example to be processed. Combustion of butane:
${C_4}{H_{10}} + {O_2} \to C{O_2} + {H_2}O$
STEP $1.$ Write all element present both sides of equation in a tabular form and write the number of atom present below them:

Elements	No of atoms on L.H.S	No of atoms on R.H.S
C	4	1
H	10	2
0	5	3

Step $2$ .Let’s start balancing from C as on LHS we have four we can do four on RHS by multiplying it with four:
H ten on LHS can be converted to ten on RHS by multiplying 5.
Step $3$ .When we have balanced C AND H the number of Oxygen changes to $13$ to balance it we multiplied $\dfrac{{13}}{2}$ on LHS.
$C = 4{\text{ C = 4}} \times {\text{1}}$
$H = 10{\text{ H = 5}} \times {\text{2}}$
$O = 2{\text{ O = 8 + 5 = 13}}$
${\text{O = 2}} \times \dfrac{{13}}{2} = 13{\text{ O = 13}}$
${\text{2}}{{\text{C}}_4}{H_{10}} + 13{O_2} \to 8C{O_2} + 10{H_2}O$
To avoid fraction terms we multiplied the whole equation by two. $3$

Note :
Remember that reactants are always written on the left side and products on the right side.
First balance elements other than hydrogen and oxygen, then after writing the equation one more time balances hydrogen and oxygen by multiplying by common factors on both sides. Our focus is to make the same numbers of atoms on both sides.