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Chapter 7
Stoichiometry of Chemical Reactions
Chapter 7 Equations and Stoichiometry
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Learning Objectives
7.1
•Derive chemical equations from narrative descriptions of chemical reactions.
•Write and balance chemical equations in molecular, total ionic, and net ionic
formats.
7.2
•Define three common types of chemical reactions (precipitation, acid-base, and
oxidation-reduction)
•Classify chemical reactions as one of these three types given appropriate
descriptions or chemical equations
•Identify common acids and bases
•Predict the solubility of common inorganic compounds by using solubility rules
•Compute the oxidation states for elements in compounds
Chapter 7 Equations and Stoichiometry
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7.3
•Explain the concept of stoichiometry as it pertains to chemical reactions
•Use balanced chemical equations to derive stoichiometric factors relating amounts
of reactants and products
•Perform stoichiometric calculations involving mass, moles, and solution molarity
7.4
•Explain the concepts of theoretical yield and limiting reactants/reagents.
•Derive the theoretical yield for a reaction under specified conditions.
•Calculate the percent yield for a reaction.
7.5
•Describe the fundamental aspects of titrations and gravimetric analysis.
•Perform stoichiometric calculations using typical titration and gravimetric data.
Chapter 7 Equations and Stoichiometry
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Chemical Equations
reactants
+
"reacts with"
products
→ "to produce”
coefficients - indicate amount of substance
Chapter 7 Equations and Stoichiometry
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Balancing Equations
Equations must be balanced.
Equal amounts of each element on each side of the equation.
4H
2O
4H, 2O
• When balancing an equation, subscripts should never change, as this changes
the chemical identity.
• Changing coefficients only changes amount of a substance, not the identity.
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Guidelines:
• write unbalanced equation
• use coefficients to indicate how many formula units are required to balance
equation
• balance those species that occur in the fewest formulas on each side.
• reduce coefficients to smallest whole number values
• When balancing reactions involving organic compounds, balance in the order: C,
H, O
KClO3
®
KCl
+
Chapter 7 Equations and Stoichiometry
O2
7
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
9
There are three major types of chemical processes occurring in aqueous
solutions:
Precipitation reactions
Acid-base reactions
Redox reactions
Chapter 7 Equations and Stoichiometry
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Precipitation Reactions (a type of double-displacement reaction)
Reactions that result in the formation of an insoluble product are known as precipitation
reactions.
A precipitate is an insoluble solid formed by a reaction in solution.
AgNO3 (aq) + KBr (aq) → AgBr(s) + KNO3 (aq)
This is an example of a metathesis reaction (also called at double replacement), a
reaction that involves the exchange of parts between the two compounds. They
exchanged anions.
Chapter 7 Equations and Stoichiometry
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Ionic Equations
Molecular equation – shows complete chemical formulas of the reactants and
products.
AgNO3(aq) + KCl(aq)→ AgCl(s) + KNO3(aq)
Complete ionic equation or ionic equation
Shows dissolved species as free ions.
Ag+(aq) + NO3–(aq) + K+ (aq) + Cl– (aq) → AgCl(s) + K+ (aq) + NO3– (aq)
Spectator ions – present but play no role in the reaction. NO3– , K+.
Chapter 7 Equations and Stoichiometry
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Net ionic equation
Shows only the species that actually take part in the reaction.
Ag+ (aq) + Cl–(aq) → AgCl(s)
1. Write a balanced molecular equation for the reaction.
2. Rewrite the equation to show the ions that form in solution when each soluble
strong electrolyte dissociates or ionizes into its component ions.
• Only dissolved strong electrolytes are written in ionic form.
Na+(aq) + Cl–(aq)
NH3(aq)
HF(aq)
3. Identify and cancel spectator ions that occur on both sides of the equation.
Chapter 7 Equations and Stoichiometry
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Solubility Guidelines for Ionic Compounds
The solubility of a substance is the amount of that substance that can be dissolved in a
given quantity of solvent.
Solubility Guidelines
A compound is probably soluble if it contains one of the following cations:
• Group 1A cation: Li+, Na+, K+, Rb+, Cs+
• Ammonium ion: NH4+
A compound is probably soluble if it contains one of the following anions:
• Halides: Cl–, Br–, I–
• Except Ag+, Hg22+, and Pb2+ compounds.
• Nitrate (NO3–), perchlorate (ClO4–), acetate (CH3CO2–),
• Sulfate (SO42–)
• Except Ba2+, Hg22+, and Pb2+.
A compound is probably insoluble if it contains one of the following anions:
• Carbonate (CO32–), phosphate (PO43–), chromate (CrO42–), sulfides (S2–)
• Except compounds containing alkali metal (Group 1A) ions and the ammonium
ion.
• Hydroxides (OH–)
• Except compounds containing alkali metal (Group 1A) ions and the Ba2+ ion.
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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From left to right: AgI, AgBr, AgCl, and AgF
Chapter 7 Equations and Stoichiometry
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Acid-Base Reactions
Acids
• Have a sour taste
• Cause litmus to change from blue to red
• React with certain metals to produce hydrogen gas
• React with carbonates and bicarbonates to produce carbon dioxide.
Bases
• Have a bitter taste
• Fell slippery. E.g.: soap
• Cause litmus to change from red to blue
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Acids are substances that are able to ionize in aqueous solutions to form a
hydrogen ion (H+) and thereby increase the concentration of H+(aq) ions.
(Arrhenius definition)
Bronsted acid - proton donor.
Monoprotic acid - each unit of acid yields one hydrogen ion.
HCl → H3O+(aq) + Cl–(aq)
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Polyprotic acids
Diprotic acid - each unit of acid gives up two H+ ions, in two separate steps.
H2SO4 → H3O+(aq) + HSO4–(aq)
HSO4–(aq) → H3O+(aq) + SO42–(aq)
Triprotic acids - yield three H+ ions
H3PO4 → H3O+(aq) + H2PO4–(aq)
H2PO4–(aq) → H3O+(aq) + HPO42–(aq)
HPO42–(aq) → H3O+(aq) + PO43–(aq)
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Bases
Bases increase the concentration of OH–(aq) ions in water. (Arrhenius definition)
Bases are substances that accept (react with) H+ ions. (Brønsted)
LiOH(s) + H2O(l) → Li+(aq) + OH–(aq)
NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)
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Strong and Weak Acids and Bases
Acids and bases that are strong electrolytes are called strong acids and strong bases.
Those that are weak electrolytes are weak acids and weak bases.
Common Strong Acids
Weak Acids are incompletely deprotonated in solution.
Ex: HF, HNO2, H2SO3, H3PO4, CH3COOH (acetic acid)
Chapter 7 Equations and Stoichiometry
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Bases
Strong bases: all Group 1A hydroxides, (LiOH, NaOH, KOH, RbOH, CsOH) and
Ba(OH)2. Ca(OH)2 and Sr(OH)2 are slightly soluble.
Be(OH)2 and Mg(OH)2 are insoluble.
Weak bases: The most common weak base is NH3. Most other weak bases
are derivatives of ammonia called amines. Ex: (CH3)3N, C5H5N, C6H5NH2.
NH3(aq) + H2O(l) → NH4+(aq) + OH–(aq)
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Neutralization Reactions and Salts
A neutralization reaction occurs when a solution of an acid and that of a base
are mixed to produce a salt (and water if the base is strong). Another type of
double displacement reaction.
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
acid
base
water
salt
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Pay attention to stoichiometry
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
2 HCl(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaCl2(aq)
H2SO4(aq) + 2 NaOH(aq) → 2 H2O(l) + Na2SO4(aq)
Weak acid and strong base
HCN(aq) + NaOH(aq) → H2O(l) +NaCN(aq)
Weak base and strong acid
HNO3(aq) + NH3(aq) → NH4NO3(aq)
In solution, NH3 reacts with water to produce NH4+ and OH–.
HNO3(aq) + NH4+(aq) + OH–(aq) → H2O(l) + NH4NO3(aq)
Chapter 7 Equations and Stoichiometry
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Oxidation –Reduction Reactions
• Loss of electrons by a substance is called oxidation.
• The gain of electrons by a substance is reduction.
• Oxidation of one substance is always accompanied by the reduction of
another as electrons are transferred between them.
LEO-GER
Loss of Electrons-Oxidation
Gain of Electrons-Reduction
OIL-RIG
Oxidation Is Loss of electrons
Reduction Is Gain of electrons
Chapter 7 Equations and Stoichiometry
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Oxidation Numbers
The oxidation number of an atom in a substance is the charge an atom would have in a
molecule (or an ionic compound) if the electrons were transferred completely.
• For an atom in its elemental form, the oxidation number is always zero.
• For any monatomic ion, the oxidation number equals the charge on the ion.
• Nonmetals usually have negative oxidation numbers, although they can sometimes be
positive.
• The oxidation number of oxygen is usually –2 in both ionic and molecular compounds. (The
major exception is in compounds called peroxides, which contain O22- ion, giving each
oxygen an oxidation number of –1.)
• The Oxidation number of hydrogen is +1 when bonded to nonmetals and –1 when bonded to
metals.
• The oxidation number of fluorine is –1 in all compounds. The other halogens have an
oxidation number of –1 in most binary compounds. When combined with oxygen, as in
oxyanions, they have positive oxidation states.
Chapter 7 Equations and Stoichiometry
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• The sum of the oxidation numbers of all atoms in a neutral
compound is zero.
• The sum of the oxidation numbers in a polyatomic ion equals
the charge of the ion.
Oxidation numbers can be determined by setting the sum of the
oxidation numbers to equal the charge on the molecule or ion.
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Maximum Oxidation number = Group number
Minimum Oxidation number = Group number – 8
–
NO3
Oxidation number of N is +5.
NH3
Oxidation number of N is –3.
N in NO3–
–
Cl in ClO3
N in NH3
Mn in KMnO4
Chapter 7 Equations and Stoichiometry
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Types of Redox Reactions
• Combination reactions
A + B → AB
C(s) + O2(g) → CO2(g)
3Mg(s) + N2(g) → Mg3N2(g)
• Decomposition Reactions
AB → A + B
2KClO3(s) → 2KCl(s) + 3O2(g)
Substances that are oxidized are reducing agents.
Substances that are reduced are oxidizing agents.
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Method of Half-Reactions
Treat oxidation and reduction as separate reactions.
C(s) + O2(g) → CO2(g)
3Mg(s) + N2(g) → Mg3N2(g)
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• Disproportionation Reaction
When one substance is oxidized and reduced in the same reaction.
Cl2 + H2O → HCl + HOCl
Hg2S → Hg(l) + HgS(s)
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Displacement Reactions
Metal displacement (Single Displacement)
Oxidation of Metals by Acids and Salts
A + BX → AX + B
Zn(s) + 2HBr(aq) → ZnBr2(aq) + H2(g)
Mn(s) + Pb(NO3)2(aq) → Mn(NO3)2(aq) + Pb(s)
These reactions are called displacement reactions because the ion in solution is
displaced or replaced through oxidation of an element.
Whenever one substance is oxidized, some other substance must be reduced.
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Half-Reactions of Displacement Reactions
Cu(s) + 2Ag+(aq) → Cu2+(aq) + 2Ag(s)
Chapter 7 Equations and Stoichiometry
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Hydrogen Displacement
Alkali metals and some alkaline earth metals (Ca, Sr, and Ba) react with water to
produce hydrogen gas.
2Na(s) + 2H2O(l) → 2NaOH(aq) + H2(g)
Some metals that don’t react with water (e.g. Fe, Zn, Mg) will react with acid to
produce hydrogen gas.
Zn(s) + 2HCl(aq) → ZnCl2(aq) + H2(g)
Halogen Displacement (the higher element ends up as the anion).
F2 > Cl2 > Br2 > I2
Cl2(g) + 2KBr(aq) → 2KCl(aq) + Br2(l)
Cl2(g) + 2KF(aq) → no reaction
Chapter 7 Equations and Stoichiometry
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Chemical Symbols on Different Levels
• Chemical symbols represent both a microscopic and macroscopic level.
2 H2(g)
+
O2(g)
®
2 H2O(g)
2 molecules
1 molecule
2 molecules
2(6x1023 molecules)
6x1023 molecules
2(6x1023 molecules)
2 moles
1 mole
2 moles
The coefficients in a balanced chemical equation can be interpreted both as the
relative numbers of molecules (or formula units) involved in the reaction and as
the relative numbers of moles.
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Stoichiometry: Chemical Arithmetic
• 2 mol H2 ≏ 1 mol O2 ≏ 2 mol H2O
• They are stoichiometrically equivalent.
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Examples
How many grams of H2O can be produced from
40.0 g O2?
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Combustion of butane, C4H10.
If we burn 1.00 g of butane, what mass of CO2 is produced?
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Reactions with Limiting Amounts of Reactants
• The reagent that is completely consumed in a reaction is called the limiting
reactant or limiting reagent, because it determines, or limits, the amount of
product formed.
• The reagent that gives the lesser amount of product is the limiting reagent.
• The amount it produces is the maximum amount of product.
• The other reactants are sometimes called excess reactants or excess reagents.
Chapter 7 Equations and Stoichiometry
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A simpler explanation.
Suppose we’re making cheese sandwiches.
The rule is one slice of cheese for two slices of bread to make
one sandwich.
1 slice cheese + 2 slices Bread → 1 Sandwich
We have six slices of cheese and twenty slices of bread. How
many sandwiches can we make?
Chapter 7 Equations and Stoichiometry
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Example
Assume that 5.52 g of sodium reacts with 5.10 g of aluminum oxide.
a) Which reactant is limiting, and which reactant is in excess?
b) How many grams of aluminum are produced?
c) What mass of excess reactant remains at the end of the reaction?
Always make sure the chemical equation is balanced!
Na(l) +
Al2O3(s) ®
Al(l) +
Chapter 7 Equations and Stoichiometry
Na2O(s)
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Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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Yields of Chemical Reactions
• The quantity of product that is calculated to form when all of the limiting
reactant reacts is called the theoretical yield. It is the maximum quantity that
can be expected on the basis of the stoichiometry of a chemical equation.
• The amount actually obtained in a reaction is called the actual yield.
• The percent yield of a reaction relates the actual yield to the theoretical
(calculated) yield:
• The percent yield is the percentage of the theoretical yield that is actually
achieved.
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Example:
Adipic acid, H2C6H8O4, is used to produce nylon. The acid is made commercially by
controlled reaction between cyclohexane (C6H12; mw = 84.14) and O2.
2 C6H12 + 5 O2 ® 2 H2C6H8O4 + 2 H2O
Assume you carry out this reaction starting with 25.0 g of cyclohexane and that
cyclohexane is the limiting reagent.
a) What is the theoretical yield of adipic acid (mw = 146.14)?
b) If you obtain 37.6 g of adipic acid from your reaction, what is the percent yield
of adipic acid?
Chapter 7 Equations and Stoichiometry
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Chapter 7 Equations and Stoichiometry
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How many moles of O₂ would be required to generate 13.0 mol of NO₂ in the
reaction below assuming the reaction has only 83.1% yield?
2 NO (g) + O₂ (g) → 2 NO₂ (g)
Chapter 7 Equations and Stoichiometry
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Learning Objectives
By the end of this section, you will be able to:
• Describe the fundamental aspects of titrations and gravimetric analysis.
• Perform stoichiometric calculations using typical titration and gravimetric data.
Chapter 7 Equations and Stoichiometry
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Gravimetric Analysis - an analytical technique based on the measurement of mass.
Ex: Calculate the concentration of an aqueous KCl solution if 25.00 mL of the
solution gives 0.430 g of AgCl when treated with excess AgNO3.
Chapter 7 Equations and Stoichiometry
Acid-Base Titrations
Titration is a procedure for determining the concentration of a solution (analyte)
using another solution of known concentration, called a standard solution.
Chapter 7 Equations and Stoichiometry
Example: 34.62 mL of 0.1510 M NaOH was needed to neutralize 50.0 mL of an
H2SO4 solution. What is the concentration of the original sulfuric acid solution?
Write a balanced equation:
Chapter 7 Equations and Stoichiometry
NaOH + HCl à
NaOH + H2SO4 à
Ba(OH)2 + HBr à
KOH + H3PO4 à
Chapter 7 Equations and Stoichiometry
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