Balancing Chemical Equations

Key Terms:
Balanced Chemical Equations
Lesson Objectives
 The student will:
 explain the roles of subscripts and coefficients in
chemical equations.
 write a balanced chemical equation when given the
unbalanced equation for any chemical reaction.
 explain the role of the law of conservation of mass in a
chemical reaction.
 balanced chemical equation
 a chemical equation in which the number of each type
of atom is equal on the two sides of the equation
 Coefficient
 a whole number that appears in front of a formula in a
balanced chemical equation
 subscript
 part of the chemical formula that indicates the number
of atoms of the preceding element
 Even though chemical compounds are broken up to
form new compounds during a chemical reaction,
atoms in the reactants do not disappear, nor do new
atoms appear to form the products. In chemical
reactions, atoms are never created or destroyed. The
same atoms that were present in the reactants are
present in the products. The atoms are merely reorganized into different arrangements. In a complete
chemical equation, the two sides of the equation must
be balanced. That is, in a balanced chemical equation,
the same number of each atom must be present on the
reactant and product sides of the equation.
Balancing Equations
 The process of writing a balanced chemical equation
involves three steps.
 Step 1: Know what the reactants and products are, and
write a word equation for the reaction.
 Step 2: Write the formulas for all the reactants and
 Step 3: Adjust the coefficients to balance the equation.
Friendly Reminder
 There are a number of elements shown in the table
that exist as diatomic molecules under normal
conditions. When any of these elements appear in
word equations, you must remember that the name
refers to the diatomic molecule and insert the diatomic
formula into the symbolic equation.
Coefficients and Subscripts
 There are two types of numbers that appear in
chemical equations. There are subscripts, which are
part of the chemical formulas of the reactants and
products, and there are coefficients that are placed in
front of the formulas to indicate how many molecules
of that substance are used or produced.
Example 1
 Write a balanced equation for the reaction that occurs
between chlorine gas and aqueous sodium bromide to
produce liquid bromine and aqueous sodium chloride.
 Step 1: Write the word equation (keeping in mind that
chlorine and bromine refer to the diatomic molecules).
 chlorine + sodium bromide yields bromine + sodium chloride
 Step 2: Substitute the correct formulas into the equation.
 Cl2+NaBr  Br2+NaCl
 Step 3: Insert coefficients where necessary to balance the
 Cl2+2 NaBr  Br2+ 2 NaCl
Example 2
 Write a balanced equation for the reaction between aluminum sulfate
and calcium bromide to produce aluminum bromide and calcium
 Step 1: Write the word equation.
 aluminum sulfate + calcium bromide yields aluminum bromide +
calcium sulfate
 Step 2: Replace the names of the substances in the word equation with
 Al2(SO4)3+CaBr2  AlBr3+CaSO4
 Step 3: Insert coefficients to balance the equation.
 products.
 Al2(SO4)3 + CaBr2  2 AlBr3+CaSO4
 Al2(SO4)3 + CaBr2  2 AlBr3+3 CaSO4
 Al2(SO4)3 + 3 CaBr2  2 AlBr3+3 CaSO4
 2 Al2(SO4)3 + 6 CaBr2  4 AlBr3 + 6 CaSO4
Practice Problem 1
 Given the following skeletal (un-balanced) equations,
balance them.
 CaCO3(s)  CaO(s)+CO2(g)
CaCO3(s)  CaO(s) + CO2(g)
 H2SO4(aq) + Al(OH)3(aq)  Al2(SO4)3(aq) + H2O(l)
3 H2SO4(aq) + 2 Al(OH)3(aq)  Al2(SO4)3(aq) + 6 H2O(l)
 Ba(NO3)2(aq) +Na2CO3(aq)  BaCO3(aq) +NaNO3(aq)
Ba(NO3)2(aq) + Na2CO3(aq)  BaCO3(aq) +2 NaNO3(aq)
Conversion of Mass in Chemical
 We already know from the law of conservation of mass
that mass is conserved in chemical reactions. But what
does this really mean? Consider the following reaction.
 Fe(NO3)3 + 3 NaOH  Fe(OH)3 + 3 NaNO3
 Verify to yourself that this equation is balanced by
counting the number of each type of atom on each
side of the equation. We can demonstrate that mass is
conserved by determining the total mass on both sides
of the equation.
Mass of Reactant Side
 Fe(NO3)3 + 3 NaOH  Fe(OH)3 + 3 NaNO3
 Mass of the Reactant Side:
 1 molecule of Fe(NO3)3 molecular weight = (1)*(241.9
daltons) = 241.9 daltons
 3 molecules of NaOHmolecular weight = (3)*(40.0
daltons) = 120. daltons
 Total mass of reactants = 241.9 daltons + 120. daltons =
361.9 daltons
Product Side Mass
 Fe(NO3)3 + 3 NaOH  Fe(OH)3 + 3 NaNO3
 1 molecule of Fe(OH)3molecular weight = (1)*(106.9
daltons) = 106.9 daltons
 3 molecules of NaNO3molecular weight = (3)*(85.0
daltons) = 255 daltons
 Total mass of products = 106.9 daltons + 255 daltons =
361.9 daltons
Lesson Summary
 Chemical equations must always be balanced.
 Balanced chemical equations have the same number
and type of each atom on both sides of the equation.
 The coefficients in a balanced equation must be the
simplest whole number ratio.
 Mass is always conserved in chemical reactions.
Practice Problem 2
 Explain in your own words why coefficients can change
but subscripts must remain constant.
 Which set of coefficients will properly balance the
following equation: C2H6 + O2  CO2 + H2O
1, 1, 1, 1
b) 1, 3, 2, 2
c) 1, 3.5, 2, 3
d) 2, 7, 4, 6
Practice Problem 3
 When properly balanced, what is the sum of all the
coefficients in the following chemical equation: SF4 +
 H2O  H2SO3 + HF
b) 7
c) 9
d) None of the above
Practice Problem 4
 When the following equation is balanced, what is the
coefficient found in front of the O2:
P4+O2+H2O  H3PO4
b) 3
c) 5
d) 7