Download Reaction Stoichiometry

Principles of Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical Quantities
and Aqueous
Reactions
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Reaction Stoichiometry
• The coefficients in a balanced chemical
equation specify the relative amounts in
moles of each of the substances involved in
the reaction.
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
Useful conversion factors are :
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
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Predicting Amounts from
Stoichiometry
• The amounts of any other substance in a
chemical reaction can be determined from the
amount of just one substance.
• How much CO2 can be made from 22.0 moles
of C8H18 in the combustion of C8H18?
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2
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1
Practice—How many moles of water are made
in the combustion of 0.10 moles of glucose?
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Given: 0.10 moles C H O
6 12 6
Find: moles H2O
mol C6H12O6
Conceptual
Plan:
Relationships:
mol H2O
1 mol C6H12O6 : 6 mol H2O
Solution:
Check:
Since there are 6× as many moles of H2O as
C6H12O6, the number makes sense.
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Example 4.1 How many grams of glucose can be
synthesized from 37.8 g of CO2 in photosynthesis?
Given: 37.8 g CO2
6 CO2 + 6 H2O → C6H12O6+ 6 O2
Conceptual g CO2
Plan:
mol CO2
mol C6H12O6
g C6H12O6
Solution:
Check: Since 6× moles of CO as C H O , but the molar mass of
2
6 12 6
C6H12O6 is 4× CO2, the number makes sense.
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Clicker # 4-1—How many grams of O2 can be made
from the decomposition of 100.0 g of PbO2?
2 PbO2(s) → 2 PbO(s) + O2(g)
[Given: Mm (PbO2)=239.2 g/mol; Mm (O2) = 32.00 g/mol]
A) 3.345 g
B) 13.38 g
C) 6.689 g
D) 50.00 g
E) 200.0 g
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2
Limiting Reactants
Explanation by Making Pizzas
• If we know that:
1 crust + 5 oz. tomato sauce + 2 cu cheese → 1 pizza
• What would happen if we had 4 crusts,
15 oz. tomato sauce and 10 cu cheese?
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Making Pizzas
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Making Pizzas
• Each ingredient could potentially make a
different number of pizzas.
• But all the ingredients have to work together!
• We have only enough tomato sauce to make 3
pizzas, so once we make 3 pizzas, the tomato
sauce runs out no matter how much of the
other ingredients we have.
Ingredient
Maximum Number of
Pizzas it can make
4 Crusts
4
15 oz. Sauce
3
10 cu Cheese
5
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3
Making Pizzas
• The tomato sauce limits the amount of pizzas
we can make. In chemical reactions we call
this the limiting reactant.
9 also known as limiting reagent
• The maximum number of pizzas we can
make depends on this ingredient. In chemical
reactions, we call this the theoretical yield.
9 It also determines the amounts of the other
ingredients we will use!
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Limiting Reactant
• For reactions with multiple reactants, it is likely that one of
the reactants will be completely used before the others.
• When this reactant is used up, the reaction stops and no
more product is made.
• The reactant that limits the amount of product is called the
limiting reactant (LR).
9 sometimes called the limiting reagent
9 the limiting reactant gets completely consumed
• Reactants not completely consumed are called excess
reactants.
• The amount of product that can be made from the limiting
reactant is called the theoretical yield (TY).
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Practice—How Many Moles of Si3N4 Can Be Made from
1.20 Moles of Si and 1.00 Moles of N2 in the Reaction
3 Si + 2 N2 → Si3N4
Given:
Find:
Conceptual
Plan:
Relationships:
1.20 mol Si, 1.00 mol N2
mol Si3N4
mol Si
mol Si3N4
mol N2
mol Si3N4
Pick least
amount
Limiting
reactant and
theoretical
yield
Solution:
Limiting
Reactant
Is Si
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Theoretical
Yield is 0.400 mol
12
4
More Making Pizzas
• Let’s now assume that as we are making
•
pizzas, we burn a pizza, drop one on the
floor, or other uncontrollable events happen
so that we make only 2 pizzas. The actual
amount of product made in a chemical
reaction is called the actual yield.
We can determine the efficiency of making
pizzas by calculating the percentage of the
maximum number of pizzas we actually
make. In chemical reactions, we call this the
percent yield (%Y).
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Example:
• When 28.6 kg of C are allowed to react with 88.2 kg
of TiO2 in the reaction below, 42.8 kg of Ti are
obtained. Find the limiting reactant, theoretical yield,
and percent yield.
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Step #1: Which is LR ?
If C is LR:
If TiO2 is LR:
smallest yield of Ti
TiO2 is limiting reactant
Step #2: calculate TY
theoretical yield
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5
Step #3: calculate percent yield (%Y)
Check the solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kg
Percent Yield = 80.9%
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Practice—How many grams of N2(g) can be made
from 9.05 g of NH3 reacting with 45.2 g of CuO ?
2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)
If 4.61 g of N2 are made, what is the percent yield?
Answer: CuO is LR; TY = 5.31 g N2 and %Y = 86.8 %
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Solutions
• When table salt is mixed with water, it seems to
disappear, or become a liquid—the mixture is
homogeneous.
9 The salt is still there, as you can tell from the taste, or
simply boiling away the water.
• Homogeneous mixtures are called solutions.
• The component of the solution that changes state is
called the solute.
• The component that keeps its state is called the
solvent.
9 If both components start in the same state, the major
component is the solvent.
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6
Solution Concentration
Chemists often use Molarity
• moles of solute per 1 liter of solution
• used because it describes how many
molecules of solute in each liter of solution
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Preparing 1 L of a 1.00 M NaCl Solution
1 liter
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Example 4.5 Find the molarity of a solution that has
25.5 g KBr dissolved in 1.75 L of solution.
Given:
Find:
25.5 g KBr, 1.75 L solution
Molarity, M
Conceptual
Plan:
Relationships:
Solution:
Check:
1 mol KBr = 119.00 g, M = moles/L
Since most solutions are between 0 and 18 M, the
answer makes sense.
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7
Clicker # 4-2 What is the molarity of a solution
containing 3.40 g of NH3 in 200.0 mL of solution?
Given: the Mm of NH3 is 17.03 g/mol
1) 0.50 M
2) 1.00M
3) 0.200 M
4) 40.0 M
5) 0.00100 M
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Example 4.6 How many liters of 0.125 M
NaOH contains 0.255 mol NaOH?
Given: 0.125 M NaOH, 0.255 mol NaOH
Find:
liters, L
Conceptual
Plan:
Relationships: 0.125 mol NaOH = 1 L solution
Solution:
Check: Since each L has only 0.125 mol NaOH, it makes sense
that 0.255 mol should require a little more than 2 L.
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Clicker # 4-3 Determine the mass of CaCl2
(Mm = 110.98) in 1.75 L of 1.50 M solution.
A) 129 g
B) 95.1 g
C) 0.237 g
D) 291 g
E) 0.291 g
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8
Practice—How would you prepare 250.0 mL
of 0.150 M CaCl2 (MM = 110.98)?
Dissolve 4.16 g of CaCl2 in enough water to total 250.0 mL
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Dilution
• Often, solutions are stored as concentrated stock
solutions.
• To make solutions of lower concentrations from
these stock solutions, more solvent is added.
9 The amount of solute doesn’t change, just the volume
of solution.
moles solute in solution 1 = moles solute in solution 2
• The concentrations and volumes of the stock and
new solutions are inversely proportional.
M1·V1 = M2·V2
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Example 4.7 To what volume should you dilute
0.200 L of 15.0 M NaOH to make 3.00 M NaOH?
Given:
Find:
V1 = 0.200 L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Conceptual
Plan:
Relationships:
M1V1 = M2V2
Solution:
Check: Since the solution is diluted by a factor of 5, the volume
should increase by a factor of 5, and it does.
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Clicker # 4-4 How would you prepare
200.0 mL of 0.25 M NaCl solution from
a 2.0 M solution?
1) Dilute 25 mL of 2.0 M solution up to
200.0 mL
2) Dilute 50 mL of 2.0 M solution up to
200.0 mL
3) Dilute 100 mL of 2.0 M solution up to
200.0 mL
4) Dilute 0.25mL of 2.0 M solution up to
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29
Solution Stoichiometry
• Since molarity relates the moles of solute to
the liters of solution, it can be used to
convert between amount of reactants and/or
products in a chemical reaction.
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10
Example 4.8 What volume of 0.150 M KCl is required to
completely react with 0.150 L of 0.175 M Pb(NO3)2 in the
2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)
Given:
Find:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
L KCl
Conceptual
Plan:
Solution:
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Practice—Solution Stoichiometry
• 43.8 mL of 0.107 M HCl is needed to
neutralize 37.6 mL of Ba(OH)2 solution.
What is the molarity of the base?
2 HCl + Ba(OH)2 → BaCl2 + 2 H2O
Answer: 0.0623 M Ba(OH)
2
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Volumetric Analysis
• An important method for determining the
amount of a particular substance is
based on measuring the volume of the
reactant solution.
9A titration is a procedure for determining the
amount of a substance, A, by adding a
carefully measured volume of a solution of
known concentration, B, until the reaction of
A and B is just complete.
9Volumetric Analysis is a method of analysis
based on titration.
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Copyright © Houghton Mifflin Company. All rights reserved.
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Figure 4.20:
Titration of an unknown amount of HCl with NaOH
(reaction completion is indicated by an indictor colour
change (i.e., clear to pink)
HCl
End-point
NaOH
NaOH=HCl
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Presentation of Lecture Outlines, 4–34
Quantitative Analysis
Volumetric Analysis
• Consider the reaction of sulfuric acid,
H2SO4, with sodium hydroxide, NaOH:
H 2SO 4 (aq ) + 2NaOH(aq ) → 2H 2O(l ) + Na 2SO 4 (aq )
– Suppose a beaker contains 35.0 mL of 0.175
M H2SO4.
– How many mL of 0.250 M NaOH must be
added to completely react with the sulfuric
acid?
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Presentation of Lecture Outlines, 4–35
Quantitative Analysis
Volumetric Analysis
• First we must convert the 0.0350 L (35.0 mL) to
moles of H2SO4 (using the molarity of the H2SO4).
• Then, convert to moles of NaOH (from the
balanced chemical equation).
• Finally, convert to volume of NaOH solution (using
the molarity of NaOH).
(0.0350L ) ×
0.175 mole H 2SO 4 2 mol NaOH 1 L NaOH soln.
×
×
=
1 L H 2SO 4 solution 1 mol H 2SO 4 0.250 mol NaOH
0.0490 L NaOH solution or
( 49.0 mL of NaOH solution)
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Presentation of Lecture Outlines, 4–36
12
Clicker # 4-5. A 25.0-mL sample of H2SO4 is neutralized
with NaOH. What is the concentration of the H2SO4 if 35.0
mL of 0.150 M NaOH are required to completely neutralize
the acid?
H2SO4(aq) + 2 NaOH (aq)→ Na2SO4 (aq)+ 2H2O(l)
a.
b.
c.
d.
e.
0.210 M
0.105 M
0.150 M
0.420 M
none of the above
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Electrolytes and Nonelectrolytes
• Materials that
•
dissolve in water to
form a solution that
will conduct
electricity are called
electrolytes.
Materials that
dissolve in water to
form a solution that
will not conduct
electricity are called
nonelectrolytes.
NaCl
Sugar
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Salt vs. Sugar Dissolved in Water
Cl- (aq)
Na+ (aq)
C6H12O6(aq)
Ionic compounds dissociate
into ions when they dissolve.
Molecular compounds do not
dissociate when they dissolve.
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13
Clicker # 4-6 Aqueous solutions of which of
the following would conduct electricity?
a.
b.
c.
d.
e.
AgNO3
C11H22O11 (sucrose: table sugar)
CH3CH2OH (ethanol)
O2
all of the above
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Also the Acids
• Acids are molecular compounds that ionize
when they dissolve in water.
9 The molecules are pulled apart by their attraction for
the water.
9 When acids ionize, they form H+ cations and anions.
• The percentage of molecules that ionize varies
from one acid to another.
• Acids that ionize virtually 100% are called
•
strong acids.
HCl(aq) → H+(aq) + Cl−(aq)
Acids that only ionize a small percentage are
called weak acids.
HF(aq) R.⇔
H+(aq) + F−(aq)
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Strong and Weak Electrolytes
• Strong electrolytes are materials that dissolve
completely as ions.
9 ionic compounds and strong acids
9 Their solutions conduct electricity well.
• Weak electrolytes are materials that dissolve
mostly as molecules, but partially as ions.
9 weak acids
9 Their solutions conduct electricity, but not well.
• When compounds containing a polyatomic ion
dissolve, the polyatomic ion stays together.
HC2H3O2(aq) ⇔ H+(aq) + C2H3O2−(aq)
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14
Clicker # 4-7
Which of the following is a weak electrolyte in
aqueous solution?
1)
H2SO3
1. HCl
2)
HNO3
2. NaCl
3)
HBr
3. H2SO4
4)
HClO4
4. HF
5) NaOH
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Dissociation and Ionization
•
•
•
When ionic compounds dissolve in water, the
anions and cations are separated from each
other. This is called dissociation.
Na2S(aq) → 2 Na+(aq) + S2–(aq)
When compounds containing polyatomic ions
dissociate, the polyatomic group stays together
as one ion.
Na2SO4(aq) → 2 Na+(aq) + SO42−(aq)
When a strong acid dissolves in water, the
molecule ionizes into H+ and anions.
H2SO4(aq) → 2 H+(aq) + SO42–(aq)
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Practice—Write the equation for the process that
occurs when the following strong electrolytes
dissolve in water.
CaCl2
CaCl2(aq) → Ca2+(aq) + 2 Cl−(aq)
HNO3
(NH4)2CO3
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15
Solubility of Ionic Compounds
• Some ionic compounds, like NaCl, dissolve very
well in water at room temperature.
• Other ionic compounds, like AgCl, dissolve hardly at
all in water at room temperature.
• Compounds that dissolve in a solvent are said to be
soluble, while those that do not are said to be
insoluble.
9 NaCl is soluble in water; AgCl is insoluble in water.
9 The degree of solubility depends on the temperature.
9 Even insoluble compounds dissolve, just not enough to
be meaningful.
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Solubility Rules
Compounds that Are Generally
Soluble in Water
Compounds Containing
the Following Ions Are
Generally Soluble
Exceptions
(when combined with ions on
the left the compound is
insoluble)
Li+, Na+, K+, NH4+
NO3 C2H3O2
–,
Cl–,
none
none
–
Br–, I–
Ag+,
SO42–
Hg2
2+,
Pb2+
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
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Group 2A
47
Solubility Rules
Compounds that Are Generally
Insoluble in Water
Compounds Containing
the Following Ions Are
Generally Insoluble
OH–
Exceptions
(when combined with ions on
the left the compound is
soluble)
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
S2–
CO32–, PO43–
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16
Determine if each of the following
is soluble in water.
KOH
KOH is soluble because it contains K+.
AgBr
AgBr is insoluble; most bromides are soluble,
but AgBr is an exception.
CaCl2
Pb(NO3)2
PbSO4
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Clicker # 4-8
Which of the following ionic compounds is
INSOLUBLE in water?
1) (NH4)2CO3
2) AgBr
3) CuSO4
4) KI
5) LiNO3
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Three Types of “chemical reactions”
to learn
1. Oxidation–Reduction Reactions
2. Acid–Base Reactions
3. Precipitation Reactions
• First are Precipitation reactions
Reactions between aqueous solutions of
ionic compounds that produce an ionic
compound that is insoluble in water are
called precipitation reactions, and the
Chem 1010 a precipitate.
insoluble productR. Helleur
is called
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17
An example
2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
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Process for Predicting the Products of
a Precipitation Reaction
1.
2.
3.
4.
5.
6.
Determine what ions each aqueous reactant has.
Determine formulas of possible products.
Determine solubility of each product in water.
If neither product will precipitate, write no
reaction after the arrow.
If either product is insoluble, write (s) after the
product that is insoluble, and (aq) after products
that remain soluble.
Balance the equation.
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Example 4.10 Write precipitation reaction between
an aqueous solution of potassium carbonate and
an aqueous solution of nickel(II) chloride.
1. Write the formulas of the reactants.
K2CO3(aq) + NiCl2(aq) →
2. Determine the possible products.
a) Determine the ions present.
(K+ + CO32–) + (Ni2+ + Cl–) →
b) Exchange the ions.
(K+ + CO32–) + (Ni2+ + Cl–) → (K+ + Cl–) + (Ni2+ + CO32–)
c) Write the formulas of the products.
K2CO3(aq) + NiCl2(aq) → KCl + NiCO3
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18
3. Determine the solubility of each product.
KCl is soluble
NiCO3 is. insoluble
4. If both products soluble, write no reaction.
does not apply since NiCO3 is insoluble
5. Write (aq) next to soluble products and (s)
next to insoluble products.
K2CO3(aq) + NiCl2(aq) → KCl(aq) + NiCO3(s)
6. Balance the equation.
K2CO3(aq) + NiCl2(aq) → 2 KCl(aq) + NiCO3(s)
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Practice—Predict the products and balance
the equation.
KCl (aq) + AgNO3(aq) →
Na2S(aq) + CaCl2(aq) →
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Ionic Equations
• Equations that describe the chemicals put into the
water and the product molecules are called
molecular equations.
2 KOH(aq) + Mg(NO3)2(aq) → 2 KNO3(aq) + Mg(OH)2(s)
• Equations that describe the actual dissolved species
are called complete ionic equations.
9 Aqueous strong electrolytes are written as ions.
9 Insoluble substances, weak electrolytes, and nonelectrolytes
are written in molecule form.
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) → 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
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19
Ionic Equations
• Ions that are both reactants and products are called
spectator ions.
2 K+(aq) + 2 OH−(aq) + Mg2+(aq) + 2 NO3−(aq) → 2 K+(aq) + 2 NO3−(aq) + Mg(OH)2(s)
•
An ionic equation in which the spectator ions are
removed is called a net ionic equation.
2 OH−(aq) + Mg2+(aq) → Mg(OH)2(s)
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Write the ionic and net ionic
equation for each.
K2SO4(aq) + 2 AgNO3(aq) → 2 KNO3(aq) + Ag2SO4(s)
2 K+(aq) + SO42−(aq) + 2 Ag+(aq) + 2 NO3−(aq)
→ 2 K+(aq) + 2 NO3−(aq) + Ag2SO4(s)
ionic
2−
+
Net ionic 2 Ag (aq) + SO4 (aq) → Ag2SO4(s)
Na2CO3(aq) + 2 HCl(aq) → 2 NaCl(aq) + CO2(g) + H2O(l)
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2. Acid–Base Reactions
• also called neutralization reactions because
the acid and base neutralize each other’s
properties
• the net ionic equation for this acid–base
reaction (strong acid and strong base) is
H+(aq) + OH−(aq) → H2O(l)
9as long as the saltR. Helleur
that Chem
forms
is soluble in water60
1010
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20
Acids and Bases in Solution
• Acids ionize in water to form H+ ions.
9 More precisely, the H from the acid molecule is donated
to a water molecule to form hydronium ion, H3O+.
• Bases dissociate in water to form OH− ions.
• In the reaction of an acid with a base, the H+ from
•
the acid combines with the OH− from the base to
make water.
The cation from the base combines with the anion
from the acid to make the salt.
acid + base → salt + water
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Example—Write the molecular, ionic, and netionic equation for the reaction of aqueous
nitric acid with aqueous calcium hydroxide.
1. Write the formulas of the reactants.
HNO3(aq) + Ca(OH)2(aq) →
2. Determine the possible products.
a) Determine the ions present when each reactant
dissociates or ionizes.
(H+ + NO3−) + (Ca2+ + OH−) →
b) Exchange the ions
(H+ + NO3−) + (Ca2+ + OH−) → (Ca2+ + NO3−) + H2O(l)
c) Write the formula of the salt.
(H+ + NO3−) + (Ca2+ + OH−) → Ca(NO3)2 + H2O(l)
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21
3. Determine the solubility of the salt.
Ca(NO3)2 is soluble
4. Write an (s) after the insoluble products and an (aq) after
the soluble products.
HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + H2O(l)
5. Balance the equation.
2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)
“molecular equation”
6. Dissociate all aqueous strong electrolytes to get
complete ionic equation.
9 not H2O
2 H+(aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) →
−
2+
“total ionic”Ca (aq) + 2 NO3 (aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic equation.
2 H+(aq) + 2 OH−(aq) → 2 H2O(l)
−(aq) → H O(l)
H+(aq) + OH
“net ionic eq.”64
2
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Practice—Predict the products and write the
balanced molecular, total ionic and net ionic
1) HCl(aq) + Ba(OH)2(aq) →
(H+ + Cl−) + (Ba2+ + OH−) → (H+ + OH−) + (Ba2+ + Cl−)
HCl(aq) + Ba(OH)2(aq) → H2O(l) + BaCl2
Molecular 2HCl(aq) + Ba(OH)2(aq) → 2 H2O(l) + BaCl2(aq)
2H+(aq) + 2Cl-(aq) + Ba2+(aq) + 2OH-(aq)→2 H2O(l) + Ba2+(aq) + 2Cl-(aq)
net ionic 2H+(aq) + 2OH- (aq)
or
→ 2 H2O (l)
H+(aq) + OH- (aq) → H2O (l)
2) H2SO4(aq) + Sr(OH)2(aq) →
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Clicker # 4-9
All the following are acid-base reactions EXCEPT
→ H2CO3(aq).
1) CO2(g) + H2O(l) ⎯⎯
→ H2SO4(aq).
2) SO3(g) + H2O(l) ⎯⎯
→ 2NaOH(aq).
3) Na2O(s) + H2O(l) ⎯⎯
→ Pb(NO3)2(aq) +
4) PbCO3(s) + 2HNO3(aq) ⎯⎯
CO2(g) + H2O(l).
→ C2H6(g).
5) C2H4(g) + H2(g) ⎯⎯
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22
Gas-Evolution Reactions
• Some reactions form a gas directly from the ion
exchange.
gas
K2S(aq) + H2SO4(aq) → K2SO4(aq) + H2S(g)
• Other reactions form a gas by the decomposition of
one of the ion exchange products into a gas and
water.
K2SO3(aq) + H2SO4(aq) → K2SO4(aq) + H2SO3(aq)
H2SO3 → H2O(l) + SO2(g)
gas
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NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
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Compounds that Undergo
Gas-Evolution Reactions
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Example 4.14 When an aqueous solution of sodium
carbonate is added to an aqueous solution of nitric
acid, a gas evolves.
1, Write the formulas of the reactants.
Na2CO3(aq) + HNO3(aq) →
2. Determine the possible products.
a) Determine the ions present when each reactant
dissociates or ionizes.
(Na+ + CO32−) + (H+ + NO3−) →
b) Exchange the anions.
(Na+ + CO32−) + (H+ + NO3−) → (Na+ + NO3−) + (H+ + CO32−)
c) Write the formula of compounds.
Na2CO3(aq) + HNO3(aq) → NaNO3 + H2CO3
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3. Check to see if either product decomposes – Yes
9
H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq) → NaNO3 + CO2(g) + H2O(l)
4. Determine the solubility of other product.
NaNO3 is soluble
5.
Balance the equation.
Na2CO3(aq) + 2 HNO3(aq) → 2 NaNO3(aq) + CO2(g) + H2O(l)
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Practice—Predict the products and
balance the equation.
HCl(aq) + Na2SO3(aq) →
(H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−)
HCl(aq) + Na2SO3(aq) → H2SO3(g) + NaCl
HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + NaCl
2 HCl(aq) + Na2SO3(aq) → H2O(l) + SO2(g) + 2 NaCl(aq)
H2SO4(aq) + CaS(aq) →
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Clicker # 4-10
The balanced net ionic equation for the reaction of
calcium carbonate with nitric acid is
→ Ca2+(aq) + 2NO2–
1) CaCO3(s) + 2HNO2(aq) ⎯⎯
(aq) + CO2(g) + H2O(l).
→ Ca2+(aq) + 2NO3–
2) CaCO3(s) + 2HNO3(aq) ⎯⎯
(aq) + CO2(g) + H2O(l).
→ Ca2+(aq) +
3) Ca(HCO3)2(s) + 2HNO3(aq) ⎯⎯
–
2NO3 (aq) + 2CO2(g) + 2H2O(l).
→
4) Ca2+(aq) + CO32–(aq) + 2H+(aq) + 2NO3–(aq) ⎯⎯
Ca(NO3)2(aq) + CO2(g) + H2O(l).
→ Ca2+(aq) + CO2(g) +
5) CaCO3(s) + 2H+(aq) ⎯⎯
H2O(l).
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3. Oxidation –Reduction reactions
• The precipitation, acid–base, and gas-
evolving reactions are all involved in
exchanging the ions in the solution.
• Other kinds of reactions involve transferring
electrons from one atom to another; these
are called oxidation–reduction reactions.
9also known as redox reactions
9many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
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Another Redox reaction: 2 Na(s) + Cl2(g) → 2 NaCl(s)
2 Na → 2 Na+ + 2 e−
Cl2 + 2 e− → 2 Cl−
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Redox Reactions of Metals with
Nonmetals
• Consider the following reactions:
•
•
4 Na(s) + O2(g) → 2 Na2O(s)
2 Na(s) + Cl2(g) → 2 NaCl(s)
The reactions involve a metal reacting with a
nonmetal.
In addition, both reactions involve the conversion
of free elements into ions.
4 Na(s) + O2(g) → 4 Na+ and 2 O2–
2 Na(s) + Cl2(g) → 2 Na+ and 2 Cl–
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Oxidation and Reduction
• In order to convert a free element into an ion,
the atoms must gain or lose electrons.
• Reactions where electrons are transferred
from one atom to another are redox reactions.
• Atoms that lose electrons are being oxidized;
atoms that gain electrons are being reduced.
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
Ger
Leo, the lion says Ger
Leo = loss of electron is oxidation
Helleur Chem 1010
Ger = gain of electron isR.reduction
Leo
77
Electron Bookkeeping
• We need a method for determining how the
electrons are transferred and how many.
• Chemists assign a number to each element in
a reaction called an oxidation state
• Even though they look like them, oxidation
states are not ion charges!
¾Oxidation states are imaginary charges assigned
based on a set of rules.
¾Ion charges are real, measurable charges.
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Rules for Assigning Oxidation States
• Rules are in order of priority.
1. Free elements have an oxidation state = 0.
9 Na = 0 and Cl2 = 0
9 Monatomic ions have an oxidation state equal to their
charge.
9 Na+ = +1 and Cl- = –1 and Mg2+= +2
2. The sum of the oxidation states of all the atoms
in a compound is 0.
9 NaCl,
(+1) + (–1) = 0
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3. The sum of the oxidation states of all the atoms in a
polyatomic ion equals the charge on the ion.
N = +5 and O = −2 in NO3–, (+5) + 3(−2) = −1
4. In their compounds, nonmetals have oxidation
states according to the table below.
Nonmetal
Oxidation State
F
−1
Example
CF4
H
+1
CH4
O
−2
CO2
Group 7A
−1
CCl4
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Example—Determine the oxidation states of
all the atoms in a propanoate ion, C3H5O2–.
• There are no free elements or free ions in
•
propanoate, so the first rule that applies is Rule 3.
(C3) + (H5) + (O2) = −1
Since all the atoms are nonmetals, the next rule we
use is Rule 4, following the elements in order:
9 H = +1
9 O = −2
(C3) + 5(+1) + 2(−2) = −1
(C3) = −2 Note: Unlike charges,
oxidation states can
C = −⅔ be fractions!
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Practice—Assign an oxidation state
to each element in the following.
• Br2
Br = 0, (Rule 1)
• K+
K = +1, (Rule 1)
• LiF
Li = +1, (Rule 2) & F = −1, (Rule 2)
• CO2
O = −2, (Rule 4) therefore C = +4
• SO42−
• Cr2O72R. Helleur Chem 1010
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Clicker # 4-11 Determine the oxidation number of the
red element in each of the following compounds:
H2PO4–
a.
b.
c.
d.
e.
+6
+5
+5
+5
+6
+6
+2
+4
+4
+4
SO32–
N2O4
+4
+4
+4
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Oxidation and Reduction
Another Definition
• Oxidation occurs when an atom’s oxidation
state increases during a reaction.
• Reduction occurs when an atom’s oxidation
state decreases during a reaction.
CH4 + 2 O2 → CO2 + 2H2O
–4 +1
0
oxidation
+4 –2
+1 –2
reduction
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Oxidation–Reduction
.
• The reactant that reduces an element in another
reactant is called the reducing agent.
9 The reducing agent contains the element that is oxidized.
• The reactant that oxidizes an element in another
reactant is called the oxidizing agent.
9 The oxidizing agent contains the element that is reduced.
2 Na(s) + Cl2(g) → 2 NaCl(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
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Example—Assign oxidation states, determine
the elements oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reaction.
Reducing
Agent
Fe +
0
Oxidizing
Agent
MnO4−
+7 −2
+ 4 H+ → Fe3+ + MnO2 + 2 H2O
+1
+3
+4 −2
+1 −2
Ger
Reduction
Oxidation
Leo
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Example—Assign oxidation states, determine
the elements oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reaction.
1. Sn4+ + Ca → Sn2+ + Ca2+
+4
0
+2
+2
Ca is oxidized, Sn4+ is reduced.
Ca is the reducing agent, Sn4+ is the oxidizing agent.
2.
F2 + S → SF4
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29
Clicker # 4-12
Which of the following conversions requires an
oxidizing agent?
→ Mn2+
1) Mn3+ ⎯⎯
→ C2H6
2) C2H4 ⎯⎯
→ SiF62–
3) SiF4 ⎯⎯
→ Cr2O72–
4) 2CrO42– ⎯⎯
→ SO3
5) SO2 ⎯⎯
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Clicker # 4-13
Which of the following reactions is an oxidation–
reduction reaction?
→ CaCl2 + H 2 O + CO 2
1) CaCO 3 + 2HCl ⎯ ⎯
→ N 2 O + 2H 2O
2) NH 4 NO 3 ⎯ ⎯
→ AgI + KNO 3
3) AgNO 3 + KI ⎯ ⎯
→ Na 2SO 4 + 2H 2 O
4) H 2 SO 4 + 2NaOH ⎯ ⎯
→ CaSO 4
5) CaO + SO 3 ⎯ ⎯
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Additional
Material
89
Oxidation-Reduction
Half-Reactions
– A half-reaction is one of the two parts of an
oxidation-reduction reaction. One involves the loss
of electrons (oxidation) and the other involves the
gain of electrons (reduction).
Fe(s ) + Cu 2+ (aq ) → Fe 2+ (aq ) + Cu(s )
–We can write this reaction in terms of two
half-reactions.
Fe(s ) → Fe 2+ (aq ) + 2e − oxidation half-reaction
+
+
Cu 2+ (aq ) + 2e − → Cu(s ) reduction half-reaction
Fe(s ) + Cu 2+ (aq
) → Fe 2+ (aq ) + Cu(s )
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Presentation of Lecture Outlines, 4–90
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Balancing Oxidation-Reduction Reactions
“using half reaction method”
• At first glance, the equation below
representing the reaction of zinc metal with
silver(I) ions might appear to be balanced.
Zn(s ) + Ag + (aq ) → Zn 2+ (aq ) + Ag(s )
total charge of +1
total charge of +2
9However, a balanced equation must have a
charge balance as well as a mass balance.
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Presentation of Lecture Outlines, 4–91
• Since the number of electrons lost in
the oxidation half-reaction must equal
the number gained in the reduction
half-reaction
Zn(s ) → Zn 2+ (aq ) + 2e −
oxidation half-reaction
2 Ag + (aq ) + 2e −
reduction half-reaction
→ 2Ag(s )
we must double the reaction involving
the reduction of the silver.
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Presentation of Lecture Outlines, 4–92
• Adding the two half-reactions together,
the electrons cancel
Zn(s ) → Zn 2+ (aq ) + 2e −
oxidation half-reaction
2 Ag + (aq ) + 2e −
reduction half-reaction
→ 2Ag(s )
Zn(s) + 2 Ag+(aq) Æ Zn2+(aq) + 2 Ag(s)
which yields the balanced oxidationreduction reaction.
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Presentation of Lecture Outlines, 4–93
31
Sample Problem
Balancing Simple Redox Equations
• Balance the following oxidation-reduction
equation using the half-reaction method:
Cr3+(aq) + Zn(s)
Cr(s) + Zn2+(aq)
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