Balancing a Chemical Reaction
Procedure
__H2 + __O2  __ H2O
_2_H2 + __O2  _ 2 _ H2O
1. Simple reactions can be balanced by inspection – using trial and error to find the
combination of stoichiometric coefficients that will balance the reaction.
More complicated reactions may require a systematic approach to balancing as follows:
methane plus oxygen react to form carbon dioxide plus water
2. Remember that you must never change the formulas of reactants or products by
changing the subscripts, only the coefficients may be changed.
3. Write out the correct formula for each reactant and write the reactant(s) on the left
side of the reaction arrow:
__CH4 + __O2 
4. Write out the correct formula for each product and write the product(s) on the right
side of the reaction arrow.
__CH4 + __O2  __ CO2  __ H2O
5. Record a list of all the elements present in the reactants below the left side of the
equation. Record a similar list of all the elements present in the products below the
right side of the equation.
__CH4 + __O2  __ CO2  __ H2O
C
H
O
C
H
O
6. Inventory the number of atoms of each element on each side of the reaction as it is
written at the present time.
__CH4 + __O2  __ CO2  __ H2O
C 1 atom
H 4 atoms
O 2 atoms
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C 1
H 2
O 3
atom
atoms
atoms
7. You can see that the numbers of H and O atoms are not balanced. Begin by balancing
the atoms in the “most complicated” molecule in the reaction. Since the carbons are
already balanced, let’s begin with the hydrogen atoms. We see that there are four
hydrogens on the left side of the reaction and two hydrogens on the right side of the
reaction. We can put a “2” as a coefficient in front of the water molecule to balance
the hydrogens:
__CH4 + __O2  __ CO2  _ 2 _ H2O
C 1 atom
H 4 atoms
O 2 atoms
C 1
H 2
O 3
atom
atoms
atoms
8. Writing the “2” in front of the water molecule changes the inventory of atoms on the
right side of the reaction, so we need to stop and take a new inventory: There is still
one carbon atom, but now there are four hydrogens and four oxygens.
__CH4 + __O2  __ CO2  _ 2 _ H2O
C 1 atom
H 4 atoms
O 2 atoms
C 1 atom
H X 4 atoms
O X 4 atoms
9. When we look at the new inventory of atoms, we see that if the only thing left to
balance is the oxygen. We can do that by placing a “2” in front of the oxygen
molecule on the left side of the equation and then taking another new inventory of
atoms.
__CH4 + _2_O2  __ CO2  _ 2 _ H2O
C 1
H 4
O X4
C 1
H X4
O X4
The reaction is now balanced.
Procedure Summary for Balancing Reactions:
1. Write the reaction placing formulas for reactants on the left and formulas for products
on the right side of the reaction..
2. Inventory the atoms on the left and right side of the reaction.
3. Beginning with an element from the “most complicated” molecule, adjust one
coefficient as needed to balance that element on both sides of the reaction, then stop
and inventory the elements again.
4. If there are other elements in the”most complicated” molecule that remain unbalanced,
balance those next, one at a time. Each time you make one change to the coefficients
of the reaction, you should stop and make a new inventory of elements.
5. Balance common small molecules such as carbon dioxide, water, or diatomic elements
last.
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Note: In a reaction in which there are polyatomic ions that do not change from the left
side of the reaction to the right side, you may balance the polyatomic ions as if they were
monatomic ions or elements, rather than balancing the individual atoms. (Both ways will
work- this way should be a little easier.)
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