Activity 151-5
Balancing Chemical Reactions
Directions: This GLA worksheet is focused on balancing chemical reactions. Part A introduces
identifying and counting atoms on both sides of a chemical reaction. Part B discusses adding coefficients
to balance the atoms. Part C discusses a shortcut that can be used when balancing double displacement
reactions. The worksheet is also accompanied by instructional videos. See
http://www.canyons.edu/Departments/CHEM/GLA for additional materials.
Part A – The Basics
According to the Law of Conservation of Matter, matter can neither be created nor destroyed. This means
that in any chemical reaction, any atoms that are present in the reactants should also be present in the
products. Subscripts in a chemical formula indicate how many of each atom is present. If there is a
subscript outside of a parenthesis, it should be multiplied by all the subscripts within the parentheses:
For example, in the compound above there are 2 aluminum atoms, 3 sulfur atoms and 12 oxygen atoms.
Example #1 Identify and count the atoms of both the reactants and products in each of the following
unbalanced chemical equations:
a) C2H6 + O2
Reactants
Atom
#
C
2
H
6
O
2
→
CO2 + H2O
Products
Atom
#
C
1
H
2
O
3
+ H2SO4 → Al2(SO4)3 +
Reactants
Products
Atom
#
Atom
#
Al
1
Al
2
H
2
H
2
S
1
S
3
O
4
O
12
b) Al
Chemistry Guided Learning Activities
Activity 151 – 5
H2
College of the Canyons
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Practice:
a) Al
+ HCl → AlCl3 + H2
Reactants
Products
Atom
#
Atom
#
_____ _____ _____ _____
_____
_____
_____
_____
_____
_____
_____
_____
b) Pb(NO3)2 + NaCl → PbCl2 + NaNO3
Reactants
Products
Atom
#
Atom
#
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
Part B – Adding coefficients to balance reactions
To balance a reaction means to make the number of atoms the same on both the reactants and products
side. To do so, coefficients need to be added to the chemical equation. Coefficients are whole numbers
that are placed in front of the element or compound in the equation to indicate how many units of each
substance participate in the chemical reaction.
When counting atoms, coefficients in front of a molecule should be multiplied by the subscripts of all
atoms in the molecule:
In the reaction below, a ‘2’ was placed in front of both the O2 and the H2O to balance the reaction. This
means that there are two O2 molecules and two H2O molecules, as shown in the diagram below. This
balances the atoms, making the number of atoms the same on both the reactants and products side. When
balancing chemical reactions, you can only change the coefficients – never change the subscripts.
Changing the subscripts changes the identity of the molecule. Coefficients of ‘1’ are omitted.
Reactants
Atom
C
H
O
Chemistry Guided Learning Activities
Activity 151 – 5
Products
#
1
4
4
Atom
C
H
O
#
1
4
4
College of the Canyons
Page 2 of 6
Balance each of the following chemical reactions by adding coefficients. It is best to add coefficients to
substances with more than one type of atom first and then add coefficients to substances with only one
type of atom last.
Example #2:
a) SiO2 +
C →
SiC +
CO
Step 1. Identify and count the number of atoms on both sides.
Reactants
Products
Atom
#
Atom
#
Si
1
Si
1
O
2
O
1
C
1
C
2
Step 2. Identify which atoms are not balanced and add coefficients, beginning with the molecule(s)
with more than one type of atom.
Since Si is already balanced, we should balance either C or O next. Since C is a single atom, we can
balance that last. To balance the O, we can add a ‘2’ to CO. This also changes the number of C, so we
then need to re-count our atoms.
SiO2 + C → SiC + 2 CO
Reactants
Products
Atom
#
Atom
#
Si
1
Si
1
O
2
O
1 2
C
1
C
2 3
Now that O is balanced, we can balance the C by adding a ‘3’ to C. We then need to re-count our
atoms.
SiO2 + 3 C → SiC + 2 CO
Reactants
Products
Atom
#
Atom
#
Si
1
Si
1
O
2
O
1 2
C
1 3
C
2 3
Our equation is balanced!
Chemistry Guided Learning Activities
Activity 151 – 5
College of the Canyons
Page 3 of 6
Practice:
a) Al
HCl → AlCl3 +
+
H2
Step 1. Identify and count the number of atoms on both sides. (See Part A).
Reactants
Products
Atom
#
Atom
#
-- 2
-- 2 _____ _____
_____ _____
_____
-- 6
_____
_____
-- 6
_____
_____
-- 6
_____
_____
-- 6
_____
Step 2. Identify which atoms are not balanced and add coefficients, beginning with the molecule(s)
with more than one type of atom.
___Al
b) C8H18
+ ___HCl → ___AlCl3 +
+
O2 →
CO2 +
___H2
H2O
Step 1. Identify and count the number of atoms on both sides.
Reactants
Products
Atom
#
Atom
#
_____
_____
_____
_____
-- 16
-- 16
_____
-- 36
_____
_____
-- 36
_____
_____
_____
-- 50
_____
-- 50
_____
Step 2. Identify which atoms are not balanced and add coefficients, beginning with the molecule(s)
with more than one type of atom.
___C8H18 +
___O2 → ___CO2 +
___H2O
Part C – Shortcut for double displacement reactions
Many reactions involve polyatomic ions that do not change from the reactants to the products. We can
count these ions as one unit so that we don’t have to count their individual atoms.
Example #3:
a) Pb(NO3)2
+
NaCl → PbCl2 + NaNO3
Step 1. Identify and count the number of atoms or polyatomic ions on both sides.
Reactants
Products
Species
#
Species
#
Pb2+
1
Pb2+
1
NO3
2
NO3
1
Na+
1
Na+
1
Cl1
Cl2
Chemistry Guided Learning Activities
Activity 151 – 5
College of the Canyons
Page 4 of 6
Step 2. Identify which atoms are not balanced and add coefficients, beginning with the molecule(s)
with more than one type of atom.
If we think of the nitrate ion as one unit, we can balance it by adding a ‘2’ to NaNO3, which also
changes the number of Na atoms.
Pb(NO3)2 + NaCl → PbCl2 + 2 NaNO3
Reactants
Products
Species
#
Species
#
Pb2+
1
Pb2+
1
NO32
NO31 2
Na+
1
Na+
1 2
Cl
1
Cl2
We can balance both Na and Cl by adding a ‘2’ to NaCl.
Pb(NO3)2 + 2 NaCl → PbCl2 + 2 NaNO3
Reactants
Products
Species
#
Species
#
Pb2+
1
Pb2+
1
NO32
NO31 2
Na+
1 2
Na+
1 2
Cl
1 2
Cl2
Our equation is balanced!
Practice:
a) HNO3 + Mg(OH)2 → H2O + Mg(NO3)2
Re-write H2O as HOH:
HNO3 + Mg(OH)2 → HOH + Mg(NO3)2
Step 1. Identify and count the number of atoms on both sides.
Reactants
Products
Species
#
Species
#
-- 2
_____
_____
_____
-- 2
_____
_____
-- 2
_____
_____
_____
_____
_____
_____
_____
_____
_____
_____
-- 2
_____
Step 2. Identify which atoms are not balanced and add coefficients, beginning with the molecule(s)
with more than one type of atom.
___HNO3 +
___Mg(OH)2 →
Chemistry Guided Learning Activities
Activity 151 – 5
___HOH + ___Mg(NO3)2
College of the Canyons
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Part D – Extra Practice
Balance each of the following chemical reactions:
a) NO2 (g) + H2 (g) → NH3 (g) + H2O (g)
b) N2H4 (l) + N2O4 (l) → N2 (g) + H2O (g)
c) C3H8 (g) + O2 (g) → CO2 (g) + H2O (g)
d) C6H14 (l) + O2 (g) → CO2 (g) + H2O (g)
e) Fe2O3(s) + CO (g) → Fe (s) + CO2 (g)
f) NO2 (g) + H2O (l) → HNO3 (aq) + NO (aq)
g) Hg2(C2H3O2)2 (aq) + KCl (aq) → Hg2Cl2 (s) + KC2H3O2 (aq)
h) H3PO4 (aq) + Ba(OH)2 (aq)
→ H2O (l) + Ba3(PO4)2 (s)
i)
Co(NO3)3 (aq) + (NH4)2S (aq) → Co2S3 (s) + NH4NO3 (aq)
j)
CO2 (g) + CaSiO3 (s) + H2O (l) → SiO2 (s) + Ca(HCO3)2 (aq)
Chemistry Guided Learning Activities
Activity 151 – 5
College of the Canyons
Page 6 of 6
Extra Practice Problems
a. 𝟐𝟐𝟐𝟐𝑶𝑶𝟐𝟐 (𝒈𝒈) + 𝟕𝟕𝑯𝑯𝟐𝟐 (𝒈𝒈) → 𝟐𝟐𝟐𝟐𝑯𝑯𝟑𝟑 (𝒈𝒈) + 𝟒𝟒𝑯𝑯𝟐𝟐 𝑶𝑶(𝒈𝒈)
b. 𝟐𝟐𝑵𝑵𝟐𝟐 𝑯𝑯𝟒𝟒 (𝒍𝒍) + 𝑵𝑵𝟐𝟐 𝑶𝑶𝟒𝟒 (𝒍𝒍) → 𝟑𝟑𝑵𝑵𝟐𝟐 (𝒈𝒈) + 𝟒𝟒𝑯𝑯𝟐𝟐 𝑶𝑶(𝒈𝒈)
c. 𝑪𝑪𝟑𝟑 𝑯𝑯𝟖𝟖 (𝒈𝒈) + 𝟓𝟓𝑶𝑶𝟐𝟐 (𝒈𝒈) → 𝟑𝟑𝟑𝟑𝑶𝑶𝟐𝟐 (𝒈𝒈) + 𝟒𝟒𝑯𝑯𝟐𝟐 𝑶𝑶(𝒈𝒈)
d. 𝟐𝟐𝑪𝑪𝟔𝟔 𝑯𝑯𝟏𝟏𝟏𝟏 (𝒍𝒍) + 𝟏𝟏𝟏𝟏𝑶𝑶𝟐𝟐 (𝒈𝒈) → 𝟏𝟏𝟏𝟏𝟏𝟏𝑶𝑶𝟐𝟐 (𝒈𝒈) + 𝟏𝟏𝟏𝟏𝑯𝑯𝟐𝟐 𝑶𝑶(𝒈𝒈)
e. 𝑭𝑭𝒆𝒆𝟐𝟐 𝑶𝑶𝟑𝟑 (𝒔𝒔) + 𝟑𝟑𝟑𝟑𝟑𝟑(𝒈𝒈) → 𝟐𝟐𝟐𝟐𝟐𝟐(𝒔𝒔) + 𝟑𝟑𝟑𝟑𝑶𝑶𝟐𝟐 (𝒈𝒈)
f. 𝟑𝟑𝟑𝟑𝑶𝑶𝟐𝟐 (𝒈𝒈) + 𝑯𝑯𝟐𝟐 𝑶𝑶(𝒍𝒍) → 𝟐𝟐𝟐𝟐𝟐𝟐𝑶𝑶𝟑𝟑 (𝒂𝒂𝒂𝒂) + 𝑵𝑵𝑵𝑵(𝒂𝒂𝒂𝒂)
g. 𝑯𝑯𝒈𝒈𝟐𝟐 (𝑪𝑪𝟐𝟐 𝑯𝑯𝟑𝟑 𝑶𝑶𝟐𝟐 )𝟐𝟐 (𝒂𝒂𝒂𝒂) + 𝟐𝟐𝟐𝟐𝟐𝟐𝟐𝟐(𝒂𝒂𝒂𝒂) → 𝑯𝑯𝒈𝒈𝟐𝟐 𝑪𝑪𝒍𝒍𝟐𝟐 (𝒔𝒔) + 𝟐𝟐𝟐𝟐𝑪𝑪𝟐𝟐 𝑯𝑯𝟑𝟑 𝑶𝑶𝟐𝟐 (𝒂𝒂𝒂𝒂)
h. 𝟐𝟐𝑯𝑯𝟑𝟑 𝑷𝑷𝑶𝑶𝟒𝟒 (𝒂𝒂𝒂𝒂) + 𝟑𝟑𝟑𝟑𝟑𝟑(𝑶𝑶𝑶𝑶)𝟐𝟐 (𝒂𝒂𝒂𝒂) → 𝟔𝟔𝑯𝑯𝟐𝟐 𝑶𝑶(𝒍𝒍) + 𝑩𝑩𝒂𝒂𝟑𝟑 (𝑷𝑷𝑶𝑶𝟒𝟒 )𝟐𝟐 (𝒔𝒔)
i. 𝟐𝟐𝟐𝟐𝟐𝟐(𝑵𝑵𝑶𝑶𝟑𝟑 )𝟑𝟑 (𝒂𝒂𝒂𝒂) + 𝟑𝟑(𝑵𝑵𝑯𝑯𝟒𝟒 )𝟐𝟐 𝑺𝑺(𝒂𝒂𝒂𝒂) → 𝑪𝑪𝒐𝒐𝟐𝟐 𝑺𝑺𝟑𝟑 (𝒔𝒔) + 𝟔𝟔𝟔𝟔𝑯𝑯𝟒𝟒 𝑵𝑵𝑶𝑶𝟑𝟑 (𝒂𝒂𝒂𝒂)
j. 𝟐𝟐𝟐𝟐𝑶𝑶𝟐𝟐 (𝒈𝒈) + 𝑪𝑪𝒂𝒂𝒂𝒂𝒂𝒂𝑶𝑶𝟑𝟑 (𝒔𝒔) + 𝑯𝑯𝟐𝟐 𝑶𝑶(𝒍𝒍) → 𝑺𝑺𝑺𝑺𝑶𝑶𝟐𝟐 (𝒔𝒔) + 𝑪𝑪𝑪𝑪(𝑯𝑯𝑯𝑯𝑶𝑶𝟑𝟑 )𝟐𝟐 (𝒂𝒂𝒂𝒂)