PREDICTING AND BALANCING CHEMICAL EQUATIONS
Chemical Equation

A chemical equation is a representation of a
chemical reaction. It is a chemist’s shorthand
description, which consists of the formulas of
substances involved in the reaction. Such
substances are the reactants, which are
conveniently written on the left side of the
equation and the products, written on the
right side, one being separated by an arrow
from the other. A chemical equation has to be
balanced in accordance with the Law of
Conservation of Mass.
Predicting Chemical Equations

Chemical reactions can be classified into four
types: Direct Combination or Synthesis,
Decomposition
or
Analysis,
Single
Replacement/Displacement or Substitution,
and Double Decomposition/Displacement. It is
possible to predict the product(s) of these
reactions by knowing the combining ability of
the different atoms or radicals. The combining
ability of the atom is indicated by the charge of
the corresponding ion.
There are some useful guidelines in writing and
predicting the outcomes of chemical reactions.
1. The products of the reaction is determined
using the following guideline:
a. Most gaseous elements such as hydrogen,
oxygen, nitrogen, chlorine, etc., exist in the
free state as diatomic molecules (H2, O2, N2,
Cl2, etc.) and should be written as such in
equations.
b. Oftentimes a radical (such as SO4 2 –, etc.) is
not changed during a chemical reaction,
hence, is carried from the left to right of the
equation without any alteration.
Examples:
6 NH4Cl + Ca3(PO4)2  2 (NH4)3PO4 + 3 CaCl2
Zn + H2SO4  ZnSO4 + H2
2. The formulas of the products must be
consistent with the rules of formula writing.
3. The type of chemical reaction may be
predicted based on the following general
patterns.
a. Direct Combination or Synthesis
In this type of reaction, two simpler substances
are combined to produce a more complex
substance in the manner: A + B  AB
Example:
2 H2 + O2  2 H2O
b. Decomposition or Analysis
This type of reaction may be taken as the
opposite of synthesis of the type: AB  A + B.
Example:
CaCO3  CaO + CO2
2 H2O  2 H2 + O2
c.
Single Replacement/Displacement or
Substitution
In this reaction, a single (free) element takes
the place of another element in a compound.
In general, if the free element is a metal, it
takes the place of the cation as in A + BC  AC
+ B and if the element is a non-metal, it takes
the place of the anion as in: A + BC  BA + C.
Examples:
Zn + 2 HCl  ZnCl2 + H2
3 Cl2 + 2 CrBr3  3 Br2 + 2 CrCl3
d. Double Decomposition/Displacement
A reaction which follows the pattern: AB + CD
 AD + CB
Example:
HCl + NaOH  NaCl + H2O
Balancing Chemical Equations
Some chemical equations are easily balanced by
inspection while others require systematic approach.
a) Inspection Method
Here are some of the guidelines in balancing
simple chemical equations by Inspection
Method:
1. Coefficients are written before the formulas.
Subscripts for atoms in the correct formulas
are never changed to balance an equation
because such change would mean another
substance or maybe wrong formula.
Begin to balance by starting with the most
complicated formula.
Example:
H3PO4 + Ba (OH)2  Ba3(PO4)2 + H2O
The most complicated formula, Ba3(PO4)2,
indicates that three barium atoms and two
phosphate ions are needed. Hence coefficient
3 is written before Ba (OH)2 and coefficient 2 is
written before H3PO4. These two coefficients
give a total of six hydrogen atoms and six
hydroxyl (OH – ) on the left so the coefficient 6
is written before H2O on the right. Thus, the
balanced equation is:
2 H3PO4 + 3 Ba(OH)2  Ba3(PO4)2 + 6 H2O
2. Coefficients are reduced to the smallest whole
number.
The equation 4 Na(s) + 2 Cl2(g)  4 NaCl(s) is
balanced but the coefficients can be divided by
two, so the final balanced equation is:
2 Na(s) + Cl2(g)  2 NaCl(s)
3. The final balanced equation should not contain
fractional coefficients (except when writing
thermochemical equations).
The equation
C2H6 + 3 ½ O2  2 CO2 + 3 H2O
is balanced but should be written as
2 C2H6 + 7 O2  4 CO2 + 6 H2O
to eliminate the fractional coefficient.
4. The balanced equations are always checked to
make sure every element is represented by the
same number of atoms on both sides of the
equation.
b) Matrix Method
Source: Alternative Chemistry Teaching Strategies
(Cabigon, 2013)
The following steps are suggested for balancing
ordinary chemical equations by Matrix Method:
1. Write the skeletal equation. The skeletal
equation should be written with the correct
formulas of reactants followed by an arrow
and the correct formulas of the products.
Remember that for most elements, the
formulas correspond to their symbols except
for seven elements which are represented as
diatomic molecules (H2, O2, N2, F2, I2, Br2, and
Cl2) and a few elements as polyatomic (P4 and
S8, although they are mostly represented as P
and S only).
2. Under the skeletal equation, draw a matrix
table with the individual elements under the
arrow column starting from the element with
the highest atomic number except oxygen
which should be written last. Indicate the
number of atoms of each element under the
compound or element column; this number is
taken from the subscripts of the elements or
from the product of the subscripts in case
there are substances enclosed in parenthesis.
See example below.
3. You may forget the given equation for now and
focus on the matrix table. Inspect each row in
the table to find out if the number of atoms is
the same before and after the arrow. Add all
the atoms on the reactant side and all the
atoms on the product side for each element or
row to check if they are the same. If not,
multiply the lesser number by a whole number
to make the number of atoms before and after
the arrow equal. The number you multiplied to
the lesser number should also be multiplied to
all the numbers in the same column. For
example, to balance the Al in the equation,
multiply 1 in Al(OH)3 column by 2 to make it
equal to 2 under Al2(SO4)3 column. You also
multiply by 2 all the numbers in the same
column as shown:
4. Go to the next atom, S. The S under H2SO4 is
multiplied by 3 to make it equal to 3 for S under
Al2(SO4)3. All the numbers in the column are
also multiplied by 3.
5. Go to the next atom, H. the total number of H
is 12 (6 under Al(OH)3 and 6 under H2SO4).
Thus, multiply 2 under H2O by 6 to obtain 12.
You also multiply all the numbers in the column
by 6.
6. Go to the last atom, O. Because the total
number for O atoms in the reactant side is 18
(6 under Al(OH)3 and 12 under H2SO4) and
equals the total number of 18 in the product
side (12 under Al2(SO4)3 and 6 under H2O), the
number of oxygen atoms is now balanced and
so the equation is also balanced. We simply use
the numbers used to multiply in each column
to write the balanced equation: 2 for Al(OH)3,
3 for H2SO4, and 6 for H2O. The balanced
equation is thus, 2 Al(OH)3 + 3 H2SO4 
Al2(SO4)3 + 6 H2O
oxidation number of zero.
Example: Na0, O20
2. Since compounds are electrically neutral, the
sum of the oxidation numbers of all the atoms
in a compound is zero.
Example:
H31+P5+O42–
(3)(1+) + (1)(5+) + (4)(2-) = 0
3. The oxidation number of a monatomic ion is
the same as the charge on the ion
(electrovalence number). In the case of
polyatomic ions, the sum of the oxidation
numbers of all the atoms is equal to the net
charge on the ion.
2.
3.
4.
5.
6.
Balancing Oxidation-Reduction Reactions
Oxidation-reduction reaction involves a change in
oxidation numbers of the atoms. Oxidation numbers
are charges assigned to the atoms of a compound
according to arbitrary rules which take into account
bond polarity.
Rules on Assigning Oxidation Numbers
1. Any uncombined atom or any atom in a
molecule of an element is assigned an
1. The oxidation number of F, the
most electronegative element, is 1in all compounds containing F.
In most oxygen-containing compounds, the
oxidation number of oxygen is 2-. In peroxides,
oxygen has an oxidation number of 1-. In OF2,
oxygen has an oxidation number of 2+.
The oxidation number of hydrogen is 1+ in all
its compounds except the metallic hydride
(CaH2 and NaH) in which hydrogen is in the 1oxidation state.
Combined alkali metals have 1+ oxidation
state.
Combined alkaline-earth metals have 2+
oxidation states.
Combined halogens have 1- oxidation state.
Steps in Balancing Oxidation-Reduction Reactions
1. Write the skeletal equation.
2. Determine the oxidation number of each
element in the equation.
3. Based on the oxidation numbers, identify the
element that undergoes oxidation (the
oxidation number increases from reactant to
product side) and the element that undergoes
reduction (the oxidation number decreases
from reactant to product side).
4. Set up the oxidation and reduction half–
reactions. Balance the elements by writing
appropriate coefficients.
5. Balance the charges on both sides of the
equation by adding enough electrons to the
more positive side.
6. Balance the number of electrons lost
(oxidation half-reaction) and gained (reduction
half-reaction) by the multiplying appropriate
factor by the half-reactions.
7. Add the two half-reactions to obtain the
partially balanced equation.
8. Use the coefficients obtained in the partially
balanced equation into the original equation
and balance the rest of the equation using the
Matrix Method.