Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 5:
Chemical Reactions
Jennifer P. Harris
Chapter 5 objectives
1. Identify the reactants and products in written reaction equations, and balance the equations by
inspection. (Section 5.1; Exercises 5.2 and 5.6)
2. Assign oxidation numbers to elements in chemical formulas, and identify the oxidizing and reducing
agents in redox reactions. (Section 5.3; only assign oxidation numbers to elements (0) and simple ions (same
as their charge)
3. Classify reactions into the categories of redox or nonredox, then into the categories of decomposition,
combination, single replacement, or double replacement. (Sections 5.4, 5.5, and 5.6; Exercise 5.20(not a,d,e)
4. Write molecular equations in total ionic and net ionic forms. (Section 5.7; Exercise 5.30 a, b, & c)
5. Classify reactions as exothermic or endothermic. (Section 5.8; Exercise 5.34)
6. Use the mole concept to do calculations based on chemical reaction equations. (Section 5.9; Exercise
5.42)
7. Use the mole concept to do calculations based on the limiting‐reactant principle. (Section 5.10;
In qualitative terms only)
8. Use the mole concept to do percentage‐yield calculations. (Section 5.11; Exercise 5.56)
CHEMICAL EQUATIONS
• Chemical equations are a convenient way to represent chemical
reactions. Chemical equations are written in terms of reactants
and products.
• A symbol is written in parentheses to the right of each reactant
and product to indicate the state or form in which the substance
exists. Gases are indicated by (g), liquids by (l), solids by (s),
and substances dissolved in water by (aq).
solid sugar = C12H22O11 (s)
and liquid water = H2O (l)
sugar dissolved in water =
C12H22O11 (aq)
CHEMICAL EQUATIONS (continued)
• REACTANTS OF A CHEMICAL EQUATION
• The reactants in a chemical equation are the
substances written on the left side of the arrow that points
toward the products. When two or more reactants are
involved in an equation, they are separated by a plus sign
(+).
Reactants
2 H2 (g) + O2 (g) → 2 H2O (l)
Products
• PRODUCTS OF A CHEMICAL EQUATION
• The products in a chemical equation are the substances
written on the right side of the arrow. When two or more
products are involved in an equation, they are separated
by a plus sign (+).
BALANCED CHEMICAL EQUATIONS
• A balanced chemical equation is one in which the number
of atoms of each element in the reactants is equal to the
number of atoms of that same element in the products.
• A reaction can be balanced by applying the law of
conservation of matter.
• Coefficients (in red below) are written to the left of each
reactant or product in order to achieve balance.
2 H2 (g) + O2 (g) → 2 H2O (l)
• Balancing chemical equations is part of the process of
stoichiometry – “weight relationships in a balanced chemical
equation”
BALANCED EQUATION PRACTICE
The process for writing and balancing chemical
equations is as follows:
1. Know (or be given) the reaction – i.e., the reactants
and products.
2. Write the correct formulas for all substances.
3. Add the additional information (g), (l), (s), (aq) – not
always necessary but good practice in that it adds
information of value
4. Balance the two sides of the equation by placing
coefficients in front of the existing formulas. If you
start with the most complicated formula, it is usually
easier.
BALANCED EQUATION PRACTICE
(continued)
• Determine if the following equations are balanced.
If not, then add coefficients to balance the
equations.
• Na2CO3 (s) → Na2O (s) + CO2 (g)
• Mg (s) + HCl (aq) → MgCl2 (aq) + H2 (g)
BALANCED EQUATION PRACTICE
(continued)
• Solid lead(II)sulfide reacts with aqueous hydrochloric
acid (HCl) to form solid lead(II) chloride and
dihydrogen sulfide gas.
• Iron(III) oxide reacts with hydrogen gas to form
elemental iron and liquid water.
• Dinitrogen pentoxide gas decomposes to form the
gases of nitrogen dioxide and oxygen.
TYPES OF CHEMICAL REACTIONS
• Chemical reactions are often classified into categories
according to characteristics of the reactions. The following is a
useful classification scheme:
REDOX REACTIONS
• The word redox is a combination of two words, reduction and oxidation.
These two words have multiple meanings when applied to chemical
reactions. Oxidation and reduction always occur together in concert.
• The concept of oxidation numbers provides a convenient way to work
with redox reactions. [we will skip the formal consideration of oxidation
number other than the oxidation number for simple ions is simply the
charge on that ion.]
OXIDIZING & REDUCING AGENTS
• In a redox reaction, the substance that contains an element
that is oxidized during the reaction is called the reducing
agent.
• In a redox reaction, the substance that contains an element
that is reduced during the reaction is called the oxidizing
agent.
REDOX REACTIONS
• Anytime an element is involved in a chemical reaction, a
redox reaction is occurring.
• We will limit our consideration of redox reactions to these
types of reactions.
• Determine the oxidation numbers (charges) on each atom
shown in the reaction. Match up the terminology with the
substances involved: oxidized/reducing agent,
reduced/oxidizing agent
• 2Ca (s) + O2(g) = 2 CaO(s)
• Al(s) + Cl2(g) = AlCl3(s) (not balanced)
• 2Na(s) +2H2O(l) = H2 (g) + 2NaOH(aq)
DECOMPOSITION REACTIONS
• In decomposition reactions, one substance is broken down
into two or more simpler substances. Decomposition
reactions may be either redox or nonredox reactions.
• The general form of the equation for a decomposition
reaction is:
A
B + C.
• An example of a redox decomposition reaction is:
2HI(g)
H2 (g) + I2 (g)
• An example of a nonredox decomposition reaction is:
H2CO3(aq)
CO2(g) + H2O(l)
DECOMPOSITION REACTIONS
(continued)
• A pictorial representation of a decomposition reaction:
2 H2O (l) → 2 H2 (g) + O2 (g)
COMBINATION REACTIONS
• In combination reactions two or more substances react to
form a single substance. Combination reactions may be
either redox or nonredox reactions.
• The general form of the equation for a combination reaction
is:
A+B
C
• An example of a redox combination reaction is:
S(s) + O2 (g)
SO2 (g)
• An example of a nonredox combination reaction is:
N2O5(g) + H2O(l)
2HNO3(aq)
COMBINATION REACTIONS
(continued)
• A pictorial representation of a combination reaction:
2 Al (s) + 3 Br2 (l) → 2 AlBr3 (s)
SINGLE-REPLACEMENT REACTIONS
• Single-replacement reactions are always redox reactions
because they occur when one element reacts with a
compound, displaces one of the elements from the
compound, and becomes part of a new compound.
• The general form of the equation for a single replacement
reaction is:
A + BX
B + AX
In this equation, A and B represent elements and AX and
BX are compounds.
• An example of a single replacement reaction is:
Zn(s) + CuSO4(aq)
Cu(s) + ZnSO4(aq)
SINGLE-REPLACEMENT REACTIONS
(continued)
• A pictorial representation of a single replacement
reaction:
Cu (s) + 2 AgNO3 (aq)→ 2 Ag (s) + Cu(NO3)2 (aq)
DOUBLE-REPLACEMENT REACTIONS
• Double-replacement reactions are never redox reactions.
These reactions often take place between substances
dissolved in water. In typical reactions, two dissolved
compounds react and exchange partners to form two new
compounds.
• The following general form of the equation for double
replacement reactions shows the partner-swapping
characteristic of the reactions:
AX + BY
BX + AY
• An example of a double-replacement reaction is:
Ba(NO3)2(aq) + Na2S(aq)
BaS(s) +2NaNO3(aq)
DOUBLE-REPLACEMENT REACTIONS
(continued)
• A pictorial representation of a double-replacement reaction:
NaCl (aq) + AgNO3 (aq)→NaNO3 (aq)
+ AgCl (s)
REACTION CLASSIFICATION
• Classify each of the following equations as redox or
nonredox reactions and as combination,
decomposition, single-replacement, or
double-replacement reactions.
• NaOH (aq) + HCl (aq) → NaCl (aq) + H2O (l)
• 2 Mg (s) + O2 (g) → 2 MgO (s)
REACTION CLASSIFICATION (continued)
• Classify each of the following equations as redox or
nonredox reactions and as combination,
decomposition, single-replacement, or doublereplacement reactions.
• Na2CO3 (s)→ Na2O (s) + CO2 (g)
• Mg (s) + 2 HCl (aq) → MgCl2 (aq) + H2 (g)
IONIC EQUATIONS
• Many reactions take place between compounds or elements
that are dissolved in water. Ionic compounds and some polar
covalent compounds break apart (dissociate) when they
dissolve in water and form ions.
• The equations for reactions that occur between dissolved
materials can be written in three ways called molecular
equations, total ionic equations, and net ionic equations.
IONIC EQUATIONS (continued)
• MOLECULAR EQUATIONS
• In a molecular equation, each compound is represented by
its formula.
• TOTAL IONIC EQUATIONS
• In a total ionic equation, all soluble ionic substances are
represented by the ions they form in solution. Substances
that do not dissolve or that dissolve but do not dissociate into
ions are represented by their formulas.
NaCl (aq) = Na+ (aq) + Cl− (aq)
Na2S (aq) = 2 Na+ (aq) + S2− (aq)
Na 3PO4 (aq) = 3 Na+ (aq) + PO43− (aq)
IONIC EQUATIONS (continued)
• NET IONIC EQUATIONS
• In a net ionic equation, only unionized or insoluble materials
and ions that undergo changes as the reaction proceeds are
represented.
• Any ions that appeared on both the left and right side of the
total ionic equation are called spectator ions and are not
included in the net ionic equation.
NaCl (aq) + AgNO3 (aq) → AgCl (s) + NaNO3 (aq)
Na+ (aq) + Ag+ (aq)
+ Na+ (aq)
→ AgCl (s)
+ Cl− (aq) + NO3− (aq)
+ NO3− (aq)
Cl− (aq) + Ag+ (aq)
→ AgCl (s)
IONIC EQUATIONS EXAMPLE
• Write the following molecular equation in total ionic and net ionic
forms. Soluble substances are indicated by (aq) after their formulas
and insoluble solids are indicated by (s) after their formulas.
BaCl2 (aq) + Na2S(aq)
BaS(s) + 2NaCl(aq)
• In total ionic form, all substances except the insoluble BaS will be
written in the form of the ions they form:
Ba2+(aq) + 2Cl-(aq)
BaS(s)
+ 2Na+(aq) + S2-(aq)
+ 2Na+(aq) + 2Cl-(aq)
• In net ionic form, all spectator ions are dropped. Both the Na+ and
Cl- ions are spectator ions because they appear on both sides of the
equation. The net ionic equation is:
Ba2+(aq) + S2-(aq)
BaS(s)
ENERGY AND REACTIONS
• In addition to changes in chemical composition, all chemical
reactions are also accompanied by changes in energy. That is,
all reactions either absorb or give up energy as they proceed.
• The energy involved in chemical reactions can take numerous
forms such as the electrical energy released by the chemical
reactions of an ordinary cell phone battery. Often, all or most of
the energy takes the form of heat.
ENERGY AND REACTIONS
(continued)
• Chemical reactions that release heat as a product are called
exothermic reactions. Ordinary combustion of a log in a
fireplace is an example of an exothermic reaction.
• While it is a physical process and not chemical, a familiar
example of an endothermic process is the melting of
ordinary ice. As the ice melts, heat is absorbed from the air
surrounding the ice.
THE MOLE AND CHEMICAL EQUATIONS
• This is the “heavy duty” side of stoichiometry.
• The mole concept can be applied to balanced chemical
equations and used to calculate mass relationships in chemical
reactions.
• Balanced equations can be interpreted in terms of the mole
concept and the results used to provide factors for use in factorunit solutions to numerical problems.
THE MOLE AND CHEMICAL EQUATIONS
EXAMPLE
• Consider the following balanced reaction equation:
2H2S(g) + 3O2(g)
2SO2 (g) + 2H2O(l)
• The mole concept can be used to write useful statements that
will be the source of factors needed to solve numerical
problems. The following are three of the possible statements:
2 mol H2S + 3 mol O2
2(6.02x1023) molecules H2S
+3(6.02x1023) molecules O2
68.2 g H2S + 96.0 g O2
2 mol SO2 + 2 mol H2O
2(6.02x1023) molecules SO2
2(6.02x1023) molecules H2O
128.2 g SO2 + 36.0 g H2O
Stoichiometry problems
• Refer to your “roadmap” – or available conversion factors
• Double check your pathway using the factor unit approach
2H2S(g) + 3O2(g)
=
2SO2 (g) + 2H2O(l)
• How many moles of oxygen are required to produce .035
moles of SO2 (g)?
• How many molecules of water would be produced by 10.0 g
of H2S?
• If 2.73 X 10-6 g of oxygen are consumed in this reaction, how
many g of water could be formed?
FACTOR-UNIT METHOD REVIEW
• Step 1: Write down the known or given quantity. Include
both the numerical value and units of the quantity.
• Step 2: Leave some working space and set the known
quantity equal to the units of the unknown quantity.
• Step 3: Multiply the known quantity by one or more factors,
such that the units of the factor cancel the units of the known
quantity and generate the units of the unknown quantity.
• Step 4: After you generate the desired units, do the
necessary arithmetic to produce the final answer.
FACTOR-UNIT METHOD EXAMPLE
• Calculate the number of moles of H2S that would need to
react with excess O2 in order to produce 115 g of SO2.
• Solution: Note that the factor used was obtained from the
two statements given earlier.
2 mol H2S
115 g SO2 
 1.79 mol H2S
128.2 g SO2
• Note that the “g SO2 ” units in the denominator of the
factor cancel the “g SO2 ” units of the known quantity, and
the “mol H2S” units of the numerator of the factor generate
the needed “mol H2S” units of the answer.
THE LIMITING REACTANT
• The limiting reactant present in a mixture of
reactants is the reactant that will run out first,
and thus, it determines the amount of product
that can be produced.
• A useful approach to solving limiting
reactant problems is to calculate the
amount of product that could be produced by
each of the quantities of reactant that are
available. The reactant that gives the least
amount of product is then the limiting
reactant.
THE LIMITING REACTANT EXAMPLE
• We will only look at the “limiting reactant” concept in a
qualitative manner
REACTION YIELDS
• The amounts of product calculated in the last three examples
are not the amounts that would be produced if the reactions
were actually done in the laboratory.
• In each case, less product would be obtained than was
calculated. There are numerous causes. Some materials
are lost during transfers from one container to another and
side reactions take place that are different from the one that
is intended to take place.
• The amount of product calculated in the examples is called
the theoretical yield. The amount of product actually
produced is called the actual yield. These two quantities are
used to calculate the percentage yield using the following
equation:
actual yield
% yield 
 100
theoretica l yeild
REACTION YIELDS EXAMPLE
• Suppose the mixture of reactants calculated earlier to give
73.7 g SO2 was done in the laboratory and only 42.7 g of SO2
was collected. What is the percentage yield of the reaction?
• Solution:
42.7 g SO2
% yield 
 100  57.9%
7.37 g SO2