CHEM 101
Principles of Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities
and Aqueous
Reactions
Quantities in Chemical Reactions
The amount of every substance used and made in a
chemical reaction is related to the amounts of all the
other substances in the reaction.
Law of Conservation of Mass
Balancing equations by balancing atoms
Study of numerical relationship between chemical
quantities in a chemical reaction is called
stoichiometry.
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
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2
Reaction Stoichiometry
Predicting Amounts from Stoichiometry
The coefficients in a balanced chemical equation specify
the relative amounts in moles of each of the substances
involved in the reaction.
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 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
2 C8H18(l) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
2 moles C8H18 : 16 moles CO2
2 moles of C8H18 react with 25 moles of O2
to form 16 moles of CO2 and 18 moles of H2O
2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O
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4
1
CHEM 101
How many moles of water are made in the
combustion of 0.10 moles of glucose?
Given: 3.4 × 1015 g C8H18
Find: g CO2
Given: 0.10 moles C6H12O6
Find: moles H2O
Conceptual
Plan:
Relationship
s:
Solution:
Check:
mol C6H12O6
Estimate the mass of CO2 produced by the
combustion of 3.5 × 1015 g petroleum.
mol H2O
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O ∴
1 mol C6H12O6 : 6 mol H2O
Conceptual g C8H18
Plan:
mol C8H18
mol CO2
g CO2
Relationships:1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01 g, 2 mol C8H18:16 mol CO2
Solution:
0.6 mol H2O = 0.60 mol H2O
Since there are 6× as many moles of H2O as
C6H12O6, the number makes sense.
Check: Since 8× moles of CO as C H , but the molar mass of
2
8 18
C8H18 is 3× CO2, the number makes sense.
5
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.
sometimes called the limiting reagent
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.
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6
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
Our balanced equation for the combustion of
methane implies that every 1 molecule of CH4
reacts with 2 molecules of O2.
H
H
C
H
H
O
+
O
O
+
O
C
O
+
H
O
O
H
+
O
H
H
8
2
CHEM 101
Limiting and Excess Reactants in the
Combustion of Methane
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)
H
If we have 5 molecules of CH4 and 8 molecules
of O2, which is the limiting reactant?
H
C
H
H
H
H
H
H
H
C
H
H
H
C
C
H
H
O
+
H
H
H
H
H
C
O
O
O
O
O
O
O
O
O
O
O
O
O
H
H
H
H
H
H
?
C
H
C
C
H
H
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
H
H
Since less CO2
can be made
from the O2
than the CH4,
the O2 is the
limiting
reactant.
H
H
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10
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?
1.20 mol Si, 1.00 mol N2
Given:
Find:
mol Si3N4
Conceptual
Plan:
H
H
H
O
O
C
mol Si
mol Si3N4
mol N2
mol Si3N4
Pick least
amount
Relationships:
Limiting
reactant and
theoretical
yield
2 mol N2 : 1 Si3N4; 3 mol Si : 1 Si3N4
Solution:
Limiting
Reactant
Theoretical and Actual Yield
we should use reaction stoichiometry to determine
the amount of product each of our reactants could
make.
The theoretical yield will always be the least possible
amount of product.
The theoretical yield will always come from the limiting
reactant.
Because of both controllable and uncontrollable
factors, the actual yield of product will always be less
than the theoretical yield.
Theoretical
Yield
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3
CHEM 101
Information
Example:
Find the limiting
Given:
28.6 kg C, 88.2 kg TiO2, 42.8
kg Ti
reactant, theoretical
yield, and percent yield. Find: Lim. Rct., Theor. Yld., % Yld.
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.
TiO2(s) + 2 C(s) →
Ti(s) + 2 CO(g)
Write a conceptual plan:
kg
C
}
smallest
amount is
from
limiting
reactant
kg
TiO2
smallest
mol Ti
13
Example:
Find the limiting
reactant, theoretical
yield, and percent
yield.
TiO2(s) + 2 C(s) →
Ti(s) + 2 CO(g)
14
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct → g rct → mol rct → mol Ti
pick smallest mol Ti → TY kg Ti → %Y Ti
Rel: 1 mol C = 12.01 g; 1 mol Ti = 47.87 g;
1 mol TiO2 = 79.87 g; 1000 g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan:
Example:
Find the limiting
reactant, theoretical
yield, and percent
yield.
TiO2(s) + 2 C(s) →
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct → g rct → mol rct → mol Ti
pick smallest mol Ti → TY kg Ti → %Y Ti
Rel: 1 mol C = 12.01 g; 1 mol Ti = 47.87 g;
1 mol TiO2 = 79.87 g; 1000 g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan:
theoretical yield
limiting reactant
smallest moles of Ti
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4
CHEM 101
Example:
Find the limiting
reactant, theoretical
yield, and percent
yield.
TiO2(s) + 2 C(s) →
Ti(s) + 2 CO(g)
Information
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find: Lim. Rct., Theor. Yld., % Yld.
CP: kg rct → g rct → mol rct → mol Ti
pick smallest mol Ti → TY kg Ti → %Y Ti
Rel: 1 mol C = 12.01 g; 1 mol Ti = 47.87 g;
1 mol TiO2 = 79.87 g; 1000 g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Apply the conceptual plan:
Solutions
When table salt is mixed with water, it seems to
disappear, or become a liquid—the mixture is
homogeneous.
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.
If both components start in the same state, the major
component is the solvent.
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Describing Solutions
Since solutions are mixtures, the composition can
vary from one sample to another.
Pure substances have constant composition.
Salt water samples from different seas or lakes have different
amounts of salt.
So to describe solutions accurately, we must describe
how much of each component is present.
We saw that with pure substances, we can describe them with a
single name because all samples are identical.
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Solution Concentration
Qualitatively, solutions are often described as dilute
or concentrated.
Dilute solutions have a small amount of solute
compared to solvent.
Concentrated solutions have a large amount of
solute compared to solvent.
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5
CHEM 101
Concentrations—Quantitative Descriptions
of Solutions
A more precise method for describing a solution is to
quantify the amount of solute in a given amount of
solution.
Solution Concentration
Molarity
moles of solute per 1 liter of solution
used because it describes how many molecules of
solute in each liter of solution
Concentration = Amount of solute in a given
amount of solution.
occasionally amount of solvent
21
Preparing 1 L of a 1.00 M NaCl Solution
22
Find the molarity of a solution that has 25.5 g KBr
dissolved in 1.75 L of solution.
Given:
Find:
Conceptual
Plan:
25.5 g KBr, 1.75 L solution
Molarity, M
g KBr
mol KBr
M
L sol’n
Relationships:
Solution:
Check:
23
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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6
CHEM 101
Example 4.6 How many liters of 0.125 M
NaOH contains 0.255 mol NaOH?
Using Molarity in Calculations
Molarity shows the relationship between the moles
of solute and liters of solution.
If a sugar solution concentration is 2.0 M, then 1
liter of solution contains 2.0 moles of sugar.
2 liters = 4.0 moles sugar
0.5 liters = 1.0 mole sugar
Given: 0.125 M NaOH, 0.255 mol NaOH
Find:
liters, L
Conceptual
Plan:
mol NaOH
L sol’n
Relationships: 0.125 mol NaOH = 1 L solution
Solution:
1 L solution : 2 moles sugar
Check:
25
Since each L has only 0.125 mol NaOH, it makes sense
that 0.255 mol should require a little more than 2 L.
26
Practice—Determine the mass of CaCl2
(MM = 110.98) in 1.75 L of 1.50 M solution.
Given:
Find:
Conceptual
Plan:
Preparing a Solution
0.150 M CaCl2, 1.75 L
mass CaCl2, g
L sol’n
mol CaCl2
g CaCl2
Relationships: 0.150 mol CaCl2 = 1 L solution; 110.98 g CaCl2 = 1 mol
Start with the amount of solution.
Use concentration as a conversion factor.
5% by mass ⇒ 5 g solute = 100 g solution
Solution:
Check:
need to know amount of solution and concentration
of solution
Calculate the mass of solute needed.
“Dissolve the grams of solute in enough solvent to
total the total amount of solution.”
Since each L has only 1.50 mol CaCl2, it makes sense
that 1.75 L should have almost 3 moles.
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7
CHEM 101
Example—How would you prepare 250.0 mL of a
1.00 M Solution of CuSO4•5 H2O (MM 249.69)?
Given: 250.0 mL solution
Find: mass CuSO4• 5 H2O, g
Conceptual mL sol’n
L sol’n
Plan:
mol CuSO4
g CuSO4
Relationship 1.00 L sol’n = 1.00 mol; 1 mL = 0.001 L; 1 mol = 249.69 g
s:
Solve:
Dissolve 62.4 g of CuSO4·5H2O in enough water to total 250.0 mL.
Check: The unit is correct; the magnitude seems
Dilution
Often, solutions are stored as concentrated stock
solutions.
To make solutions of lower concentrations from
these stock solutions, more solvent is added.
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
reasonable as the volume is ¼ of a liter.
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30
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:
Conceptual
Plan:
Relationships:
V1 = 0.200 L, M1 = 15.0 M, M2 = 3.00 M
V2, L
V1, M1, M2
V2
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.
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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8
CHEM 101
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
reaction 2 KCl(aq) + Pb(NO3)2(aq) → PbCl2(s) + 2 KNO3(aq)
Given:
Find:
Conceptual
Plan:
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?
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
L KCl
L Pb(NO3)2
mol Pb(NO3)2
Practice—Solution Stoichiometry
mol KCl
L KCl
2 HCl + Ba(OH)2 → BaCl2 + 2 H2O
Relationships: 1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 : 2 mol KCl
Solution:
Check:
Since we need 2× moles of KCl as Pb(NO3)2, and the
molarity of Pb(NO3)2 > KCl, the volume of KCl should be
more than 2× the volume of Pb(NO3)2.
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34
Practice—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(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(aq)
Given: 0.0438
43.8 mL
L of 0.107 M HCl, 0.0376
37.6 mLLBa(OH)
Ba(OH)22
MBa(OH)
Ba(OH)22
Find: M
Conceptual
Plan:
L HCl
mol HCl
mol Ba(OH)2
M Ba(OH)2
L Ba(OH)2
Relationships: 1 mL = 0.001 L, 1 L HCl = 0.107 mol, 1 mol Ba(OH)2 : 2 mol HCl
Solution:
Check:
What happens when a solute dissolves?
There are attractive forces between the solute particles
holding them together.
There are also attractive forces between the solvent
molecules.
When we mix the solute with the solvent, there are
attractive forces between the solute particles and the
solvent molecules.
If the attractions between solute and solvent are strong
enough, the solute will dissolve.
The units are correct; the magnitude makes sense.
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9
CHEM 101
Table Salt Dissolving in Water
Each ion is attracted
to the surrounding
water molecules and
pulled off and away
from the crystal.
When it enters the
solution, the ion is
surrounded by water
molecules, insulating
it from other ions.
37
The result is a
solution with freemoving charged
particles able to
conduct electricity.
Salt vs. Sugar Dissolved in Water
In order to conduct electricity, a material must
have charged particles that are able to flow.
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.
38
Acids
Acids are molecular compounds that ionize when
they dissolve in water.
The molecules are pulled apart by their attraction for the
water.
When acids ionize, they form H+ cations and anions.
Ionic compounds dissociate
into ions when they dissolve.
Molecular compounds do not
dissociate when they dissolve.
39
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) ⇔ H+(aq) + F−(aq)
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CHEM 101
Strong and Weak Electrolytes
Strong electrolytes are materials that dissolve
completely as ions.
ionic compounds and strong acids
Their solutions conduct electricity well.
Weak electrolytes are materials that dissolve
mostly as molecules, but partially as ions.
weak acids
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)
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.
NaCl is soluble in water; AgCl is insoluble in water.
The degree of solubility depends on the temperature.
Even insoluble compounds dissolve, just not enough to be
meaningful.
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42
Solubility Rules
Compounds that Are Generally Soluble in Water
Compounds Containing
the Following Ions Are
Generally Soluble
Li+,
Na+,
K +,
Exceptions
(when combined with ions on
the left the compound is
insoluble)
NH4+
none
NO3–, C2H3O2–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
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Solubility Rules
Compounds that Are Generally Insoluble in Water
Compounds Containing
the Following Ions Are
Generally Insoluble
OH–
S2–
CO32–, PO43–
Exceptions
(when combined with ions on
the left the compound is
soluble or slightly soluble)
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
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11
CHEM 101
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
CaCl2 is soluble; most chlorides are soluble,
and CaCl2 is not an exception.
Pb(NO3)2
Pb(NO3)2 is soluble because it contains NO3−.
PbSO4
PbSO4 is insoluble; most sulfates are soluble,
but PbSO4 is an exception.
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 insoluble product is
called a precipitate.
45
Write the equation for the precipitation
reaction between an aqueous solution of
potassium carbonate and an aqueous solution
of nickel(II) chloride.
1.
3.
4.
5.
Determine the ions present.
(K+ + CO32–) + (Ni2+ + Cl–) →
Exchange the ions.
(K+ + CO32–) + (Ni2+ + Cl–) → (K+ + Cl–) + (Ni2+ + CO32–)
c)
Write the formulas of the products.
b)
If both products soluble, write no reaction.
does not apply since NiCO3 is insoluble
Determine the possible products.
a)
Determine the solubility of each product.
KCl is soluble
NiCO3 is insoluble
Write the formulas of the reactants.
K2CO3(aq) + NiCl2(aq) →
2.
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6.
Write (aq) next to soluble products and (s) next to
insoluble products.
K2CO3(aq) + NiCl2(aq) → KCl(aq) + NiCO3(s)
Balance the equation.
K2CO3(aq) + NiCl2(aq) → 2 KCl(aq) + NiCO3(s)
cross charges and reduce
K2CO3(aq) + NiCl2(aq) → KCl + NiCO3
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12
CHEM 101
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.
Aqueous strong electrolytes are written as ions.
Insoluble substances, weak electrolytes, and nonelectrolytes
are written in molecule form.
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)
solids, liquids, and gases are not dissolved, therefore, 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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Acid–Base Reactions
50
Common Acids
Also called neutralization reactions because the acid
and base neutralize each other’s properties
2 HNO3(aq) + Ca(OH)2(aq) → Ca(NO3)2(aq) + 2 H2O(l)
Acids ionize in water to form H+ ions.
Bases dissociate in water to form OH− ions.
the net ionic equation for an acid–base reaction is
H+(aq) + OH−(aq) → H2O(l)
as long as the salt that forms is soluble in 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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13
CHEM 101
Common Bases
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
53
Write the molecular, ionic, and net-ionic
equation for the reaction of HNO3 (aq)
with Ca(OH)2 (aq).
Write the formulas of the reactants.
HNO3(aq) + Ca(OH)2(aq) →
1.
Determine the possible products.
2.
a)
b)
c)
Determine the ions present when each reactant dissociates or
ionizes.
(H+ + NO3−) + (Ca2+ + OH−) →
Exchange the ions, H+ combines with OH− to make H2O(l).
(H+ + NO3−) + (Ca2+ + OH−) → (Ca2+ + NO3−) + H2O(l)
Write the formula of the salt.
Cross the charges.
(H+ + NO3−) + (Ca2+ + OH−) → Ca(NO3)2 + H2O(l)
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54
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)
3.
Dissociate all aqueous strong electrolytes to get complete
ionic equation.
+
2 H (aq) + 2 NO3−(aq) + Ca2+(aq) + 2 OH−(aq) →
Ca2+(aq) + 2 NO3−(aq) + H2O(l)
7.
Eliminate spectator ions to get net-ionic equation.
2 H+(aq) + 2 OH−(aq) → 2 H2O(l)
H+(aq) + OH−(aq) → H2O(l)
6.
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14
CHEM 101
Gas-Evolution Reactions
Some reactions form a gas directly from the ion
exchange.
NaHCO3(aq) + HCl(aq) → NaCl(aq) + CO2(g) + H2O(l)
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)
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58
Compounds that Undergo
Gas-Evolution Reactions
Predict the products and balance the
equation.
HCl(aq) + Na2SO3(aq) →
(H+ + Cl−) + (Na+ + SO32−) → (H+ + SO32−) + (Na+ + Cl−)
Reactant
Type
Reacting
Ion
Decom
Exchange -pose?
With
Product
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2 HCl(aq) →
2 KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2 HCl(aq) →
2 KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2 HCl(aq) →
2 KCl(aq) + SO2(g) + H2O(l)
(NH4)nanion
base
NH4OH
yes
NH3
KOH(aq) + NH4Cl(aq) →
KCl(aq) + NH3(g) + H2O(l)
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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) →
(H+ + SO42−) + (Ca2+ + S2−) → (H+ + S2−) + (Ca2+ + SO42−)
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4
H2SO4(aq) + CaS(aq) → H2S(g) + CaSO4(s)
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15
CHEM 101
Redox without Combustion
2 Na(s) + Cl2(g) → 2 NaCl(s)
Other Patterns in 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.
also known as redox reactions
many involve the reaction of a substance with O2(g)
4 Fe(s) + 3 O2(g) → 2 Fe2O3(s)
2 Na → 2 Na+ + 2 e−
Cl2 + 2 e− → 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.
Of course, if one atom loses electrons, another must
accept them.
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.
Ger
Na+Cl–(s)
2 Na(s) + Cl2(g) → 2
Na → Na+ + 1 e– oxidation
Cl2 + 2 e– → 2 Cl– reduction
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Leo
Electron Bookkeeping
For reactions that are not metal + nonmetal, or
do not involve O2, we need a method for
determining how the electrons are transferred.
Chemists assign a number to each element in a
reaction called an oxidation state that allows
them to determine the electron flow in the
reaction.
Even though they look like them, oxidation states
are not ion charges!
Oxidation states are imaginary charges assigned based
on a set of rules.
Ion charges are real, measurable charges.
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CHEM 101
Rules for Assigning Oxidation States
1.
Rules are in order of priority.
Free elements have an oxidation state = 0.
Rules for Assigning Oxidation States
3.
Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)
2.
Monatomic ions have an oxidation state equal to
their charge.
Na = +1 and Cl = –1 in NaCl
3.
(a) The sum of the oxidation states of all the
atoms in a compound is 0.
Na = +1 and Cl = –1 in NaCl, (+1) + (–1) = 0
(b) 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.
(a) Group I metals have an oxidation state of
+1 in all their compounds.
Na = +1 in NaCl
(b) Group II metals have an oxidation state of
+2 in all their compounds.
Mg = +2 in MgCl2
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Rules for Assigning Oxidation States
5.
In their compounds, nonmetals have oxidation
states according to the table below.
Nonmetals higher on the table take priority.
Nonmetal
Oxidation State
Example
F
−1
CF4
H
+1
CH4
O
−2
CO2
Group 7A
−1
CCl4
Group 6A
−2
CS2
−3
NH3
Group 5A
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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 3b.
(C3) + (H5) + (O2) = −1
Since all the atoms are nonmetals, the next rule we
use is Rule 5, following the elements in order:
H = +1
O = −2
(C3) + 5(+1) + 2(−2) = −1
(C3) = −2 Note: Unlike charges,
oxidation states can
C = −⅔
be fractions!
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CHEM 101
Practice—Assign an oxidation state to
each element in the following.
• Br2
Br = 0, (Rule 1)
• K+
K = +1, (Rule 2)
• LiF
Li = +1, (Rule 4a) & F = −1, (Rule 5)
• CO2
O = −2, (Rule 5) & C = +4, (Rule 3a)
• SO42−
O = −2, (Rule 5) & S = +6, (Rule 3b)
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 + 2 H2O
–4 +1
0
+4 –2
+1 –2
oxidation
• Na2O2 Na = +1, (Rule 4a) & O = −1 , (Rule 3a)
reduction
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Oxidation–Reduction
Oxidation and reduction must occur simultaneously.
If an atom loses electrons another atom must take them.
The reactant that reduces an element in another
reactant is called the reducing agent.
The reducing agent contains the element that is oxidized.
The reactant that oxidizes an element in another
reactant is called the oxidizing agent.
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Assign oxidation states, determine the
elements oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reaction.
Reducing
Agent
Oxidizing
Agent
Fe + MnO4− + 4 H+ → Fe3+ + MnO2 + 2 H2O
0
+7
−2
+1
+3
+4 −2
+1 −2
The oxidizing agent contains the element that is reduced.
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
Na is oxidized, Cl is reduced
Na is the reducing agent, Cl2 is the oxidizing agent
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Reduction
Oxidation
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CHEM 101
Assign oxidation states, determine the
elements oxidized and reduced, and
determine the oxidizing agent and reducing
agent in the following reaction.
Sn4+ + Ca → Sn2+ + Ca2+
+4
0
+2
+2
Ca is oxidized, Sn4+ is reduced.
Ca is the reducing agent, Sn4+ is the oxidizing agent.
Combustion Reactions
Reactions in which O2(g) is a reactant are called
combustion reactions.
Combustion reactions release lots of energy.
Combustion reactions are a subclass of oxidation–
reduction reactions.
F2 + S → SF4
2 C8H18(g) + 25 O2(g) → 16 CO2(g) + 18 H2O(g)
0
0
+4−1
S is oxidized, F is reduced.
S is the reducing agent, F2 is the oxidizing agent.
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Combustion Products
To predict the products of a combustion reaction,
combine each element in the other reactant with
oxygen.
Reactant
Combustion
Product
contains C
CO2(g)
contains H
H2O(g)
contains S
SO2(g)
contains N
NO(g) or NO2(g)
contains metal
M2On(s)
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Write the equation for the complete
combustion 76
of CH3OH(l).
1.
Write the reactants.
combustion is reacting with O2(g)
2.
CH3OH(l) + O2(g) →
Combine O with each element to make a product.
C + O make CO2(g); H + O make H2O(g)
3.
CH3OH(l) + O2(g) → CO2(g) + H2O(g)
Balance the equation.
2 CH3OH(l) + 3 O2(g) → 2 CO2(g) + 4 H2O(g)
Tro, Principles of Chemistry: A Molecular Approach
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