Balancing Chemical Equation
Deo Kumar Kharka
Tashi Tenzin
Shankarlal Chettri
Ugyen Thinley
What is chemical equation?
 Chemical reaction equations are mathematical and symbolic
models of real world compound and events.
When a chemical reaction occurs, it can be described by an equation.
This shows the chemicals that react (called the reactants) on the lefthand side,
The chemicals that they produce (called the products) on the righthand side.
The chemicals can be represented by their names or by their chemical
Unlike mathematical equations, the two sides are separated by an
arrow, that indicates that the reactants form the products and not the
other way round.
There could be two kinds of equation:
1. skeleton equation - is an equation which just represents
a chemical change but it is not – balanced.
Eg. KNO3
KNO2 + O2
2. balanced equation –is the equation in which the total
number of atoms in the reactant is same as the number of
the atom in the product formed.
Eg. 2KNO3
2KNO2 + O2
How to balance the chemical equation
1. Count the number of times (frequency) an element is
occurring on the both the sides.
2. an element with less frequency is balanced first.
3. when two or more elements have same frequency then
metallic element is preferred.
1. Inspection/hit and trail method
2. Algebraic method
3. Partial equation method
Hit and Trial method
Right hand side
N2O + H2O
left hand side
N=2 (balanced)
O=2 (not balanced)
H=2 (not balanced)
O and H can be balance by adding coefficient 2 to the H2O
N2O +2 H2O
Algebraic method
 EG. C4H10 + O2
CO2 + H20
 put unknown coefficients in front of each molecular
species in the equation:
xC4H10 + yO2
zCO2 + wH20
write down the balance conditions for each element in
terms of the unknowns (x,y,z,w).
 Carbon balance:
4x= z
 Hydrogen balance condition is:
 oxygen balance condition gives:
2y= 2z + w
 We should recognize that chemical equations specify relative amounts of
reactants and products. Thus, one of the coefficients is arbitrary.
 we may take one of the coefficients to be 1. (according to convenience)
o Assume x= 1,
4x= z therefore, z= 4
10x=2w therefore, w= 5
 from oxygen balance condition, y= 2z + w
2y = 2 x 4 + 5
y= 13/2
 Thus, the balanced equation could be written as :
C4H10 +
4CO2 + 5H20
 Multiply through the equation by 2 on both sides (it can be
treated as an algebraic equation in this way)
C4H10 + 13/2 O2
4CO2 + 5H20
2C4H10 + 13 O2
8CO2 + 10H20
Partial Equation Method
 When equations contain many reactants and products they
cannot be balanced by the hit and trial method.
 They are then balanced by the partial equation method.
 In this method the overall reaction is assumed to take place
through two or more simpler reactions, which can be
represented by partial equations.
Split the chemical equation into two or more partial
Na OH + Cl2
NaCl +NaClO3 + H2O
Partial eq.1
Na OH + Cl2
NaCl +NaClO + H2O
Partial eq.2
NaClO3 +NaCl
ii. Each partial equation is separately balanced by the hit and trial
Balanced Partial eq.1
2NaOH + Cl2
Balanced Partial eq.2
3 NaClO
NaCl + NaClO + H2O
NaClO3 + 2NaCl
 These balanced partial equations are multiplied with suitable
coefficients in order to exactly cancel out those common
substances which do not appear in the overall chemical
2NaOH + Cl2
NaCl + NaClO + H2O….. X 3
 The balanced partial equations so obtained, are added to
arrive at the balanced chemical equation.
6NaOH +3 Cl2
3NaCl + 3 NaClO + 3H2O
3 NaClO
6NaOH +3Cl2
NaClO3 + 2NaCl
5NaCl + NaClO3 + H2O

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