Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 4
Chemical
Quantities and
Aqueous
Reactions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Reaction Stoichiometry
• the numerical relationships between chemical amounts
•
in a reaction is called 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 molecules of C8H18 react with 25 molecules of O2
to form 16 molecules of CO2 and 18 molecules of H2O
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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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
22.0 moles C8 H18 
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16 mol CO2
2 mol C8 H18
 176 moles CO2
3
Example – Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
• assuming that gasoline is octane, C8H18, the equation
•
for the reaction is:
2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
the equation for the reaction gives the mole relationship
between amount of C8H18 and CO2, but we need to
know the mass relationship, so the Concept Plan will
be:
g C8H18
mol C8H18
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mol CO2
g CO2
4
Example – Estimate the mass of CO2 produced in
2004 by the combustion of 3.4 x 1015 g gasoline
Given:
Find:
3.4 x 1015 g C8H18
g CO2
Concept Plan:
g C8H18
mol C8H18
1 mol
114.22 g
mol CO2
1 6 mol CO
g CO2
44.01 g
2
1 mol
2 mol C 8 H 18
Relationships: 1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2
Solution:
3 .4  10
15
 1.0  10
g C 8 H 18 
16
g CO
1 mol C 8 H 18
114.22 g C 8 H 18

16 mol CO
2

2 mol C 8 H 18
44.01 g CO
1 mol CO
2
2
2
Check: since 8x moles of CO as C H , but the molar mass of C H is
2
8 18
8 18
3x CO2, the number makes sense
Practice
• According to the following equation, how
many milliliters of water are made in the
combustion of 9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
1.
2.
3.
4.
convert 9.0 g of glucose into moles (MM 180)
convert moles of glucose into moles of water
convert moles of water into grams (MM 18.02)
convert grams of water into mL
a) How? what is the relationship between mass and
volume?
density of water = 1.00 g/mL
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Practice
According to the following equation, how many
milliliters of water are made in the combustion of
9.0 g of glucose?
C6H12O6(s) + 6 O2(g)  6 CO2(g) + 6 H2O(l)
9.0 g C 6 H12O 6 x
1 mole C 6 H12O 6
2
1.80 x 10 g
x
6 mole H 2 O
1 mole C 6 H12O 6
x
18.0 g H 2 O
1 mole H 2 O
x
1 mL H 2 O
1.00 g H 2 O
 5.4 mL H 2 O
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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
 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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Things Don’t Always Go as Planned!
• many things can happen during the course of an
experiment that cause the loss of product
• the amount of product that is made in a reaction
is called the actual yield
generally less than the theoretical yield, never more!
• the efficiency of product recovery is generally
given as the percent yield
Percent Yield 
actual yield
100%
theoretica l yield
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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
+
H
H
+
O
O
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O
O
H
H
10
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
• 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
H
H
H
H
H
C
+
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
O
?
H
H
11
Limiting and Excess Reactants in the
Combustion of Methane
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
H
C
H
H
H
H
H
H
H
C
H
H
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
8 molecules CH 4 
10 molecules O 2 
2 molecules CO2
1 molecules CH 4
2 molecules CO 2
2 molecules O 2
since less CO2
can be made
from the O2 than
the CH4, the O2
is the limiting
reactant
 16 molecules CO 2
 10 molecules CO2
12
Example 4.4
Finding Limiting Reactant,
Theoretical Yield, and
Percent Yield
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.
TiO 2 (s)  2 C (s)  Ti (s)  2 CO (g)
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Example:
When 28.6 kg of C reacts with 88.2
kg of TiO2, 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 down the given quantity and its units.
Given:
28.6 kg C
88.2 kg TiO2
42.8 kg Ti produced
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Information
Example:
Given: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti
Find the Limiting
Reactant, Theoretical
Yield, and Percent Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Write down the quantity to find and/or its units.
Find: limiting reactant
theoretical yield
percent yield
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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.
• Write a Concept Plan:
kg
C
1000 g
g
C
1 kg
kg
TiO2
1000 g
g
TiO2
47.87 g
1 mol Ti
1 mol TiO 2
mol
TiO2
79.87 g TiO 2
g Ti
1 kg
1 000 g
mol
Ti
1 mol Ti
2 mol C
12.01 g C
1 kg
smallest
mol Ti
1 mol C
mol
C
mol
Ti
1 mol Ti
}
smallest
amount is
from
limiting
reactant
1 mol TiO 2
kg Ti
T.Y.
% yield 
act. yield
theor. yield
% Yield
17
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
• Collect Needed Relationships:
1000 g = 1 kg
Molar Mass TiO2 = 79.87 g/mol
Molar Mass Ti = 47.87 g/mol
Molar Mass C = 12.01 g/mol
1 mole TiO2 : 1 mol Ti (from the chem. equation)
2 mole C  1 mol Ti (from the chem. equation)
18
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=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Apply the Concept Plan:
28.6 kg C 
1000 g

1 kg
88.2 kg TiO 2 
1000 g
1 kg
1 mole C

12.01 g C

1 mol Ti
2 mol C
1 mole TiO 2
79.87 g TiO 2
Limiting Reactant
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3
 1.1907 10 mol Ti

1 mol Ti
3
 1.1043 10 mol Ti
1 mol TiO 2
smallest moles of Ti
19
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=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Apply the Concept Plan:
3
1.1 0 43  10 mol Ti 
47.87 g Ti
1 mol

1 kg
 52.9 kg Ti
1000 g
Theoretical Yield
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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=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
Example:
Find the Limiting
Reactant, Theoretical
Yield, and Percent
Yield.
TiO2(s) + 2 C(s) 
Ti(s) + 2 CO(g)
• Apply the Concept Plan:
Actual Yield
Theoretica
 100 %  Percent Yield
l Yield
42.8 kg Ti
 100 %  8 0 . 9 %
52.9 kg Ti
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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=12.01g; 1 mol Ti =47.87g;
1 mol TiO2 = 79.87g; 1000g = 1 kg;
1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti
• Check the Solutions:
Limiting Reactant = TiO2
Theoretical Yield = 52.9 kg
Percent Yield = 80.9%
Since Ti has lower molar mass than TiO2, the T.Y. makes sense
The Percent Yield makes sense as it is less than 100%.
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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)
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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)
Given:
Find:
9.05 g NH3, 45.2 g CuO
g N2
Concept Plan: g NH
3
mol NH3
mol N2
1 mol
1 mol N 2
17.03 g
2 mol NH 3
g CuO
mol CuO
mol N2
1 mol
1 mol N 2
79.55 g
3 mol CuO
smallest moles N2
1 mol
g N2
28.02 g
Relationships:
1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g
2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2
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)
Solution:
9 .05 g NH 3 
4 5.2 g CuO 
1 mol NH 3

17.03 g NH 3
2 mol NH 3
1 mol CuO
1 mol N 2

79.55 g CuO
0 .189 mol N 2 
Check:
1 mol N 2
3 mol CuO
28.02 g N 2
1 mol N 2
 0 .266 mol N 2
 0 .189 mol N 2
 5 .30 g N 2
units are correct, and since there are fewer moles
of N2 than CuO in the reaction and N2 has a
smaller mass, the number makes sense
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 identical
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27
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
quantitatively, the relative
amount of solute in the solution
is called the concentration
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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
molarity, M 
amount of solute (in moles)
amount of solution (in L)
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Preparing 1 L of a 1.00 M NaCl Solution
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30
Example 4.5 – Find the molarity of a solution that
has 25.5 g KBr dissolved in 1.75 L of solution
•
Sort
Information
•
Strategize
Given:
Find:
25.5 g KBr, 1.75 L solution
Molarity, M
Concept Plan:
g KBr
mol KBr
M 
1 mol
mol
L
M
L sol’n
1 mol KBr = 119.00 g,
M = moles/L
119.00 g
Relationships:
•
•
Follow the
Concept Plan
to Solve the
problem
Check
Solution:
1 mol KBr
2 5.5 g KBr 
 0.21 4 29 mol KBr
119.00 g KBr
molarity,
M 
moles KBr
L solution

0 . 21 4 29 mol KBr
 0.122 M
1.75 L
Check: since most solutions are between 0 and
18 M, the answer makes sense
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
• 1 L solution : 2 moles sugar
2 mol sugar
1 L solution
1 L solution
2 mol sugar
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32
Example 4.6 – How many liters of 0.125 M NaOH
contains 0.255 mol NaOH?
•
Sort
Information
•
Strategize
Given: 0.125 M NaOH, 0.255 mol NaOH
Find:
liters, L
Concept Plan:
L sol’n
mol NaOH
1 L solution
0.125 mol NaOH
Relationships:
•
•
Follow the
Concept Plan
to Solve the
problem
Check
0.125 mol NaOH = 1 L solution
Solution:
0 . 2 55 mol NaOH 
1 L solution
 2.04 L solution
0.125 mol NaOH
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
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
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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?
•
Sort
Information
•
Strategize
Given: V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 M
V2, L
Find:
Concept Plan:
V1, M1, M2
V2
M 1  V1
M2
Relationships:
•
Follow the
Concept Plan
to Solve the
problem
•
Check
Solution:
 V2
M1V1 = M2V2
mol 

 15.0
  0 . 200 L 
L


 1.00 L
mol 

3.00


L


Check: since the solution is diluted by a factor
of 5, the volume should increase by a
factor of 5, and it does
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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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)
•
•
Sort
Information
Strategize
Given:
0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2
Find:
L KCl
Concept Plan:
L Pb(NO3)2
mol Pb(NO3)2
0.175 mol
1 L Pb(NO
Relationships:
•
•
Follow the
Concept
Plan to
Solve the
problem
Check
mol KCl
1 L KCl
2 mol KCl
3 )2
1 mol Pb(NO
L KCl
3 )2
0.150 mol
1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol,
1 mol Pb(NO3)2 = 2 mol KCl
Solution:
0 .150 L Pb(NO
3 )2 
0.175 mol
1 L Pb(NO
3 )2

2 mol KCl
1 mol Pb(NO

3 )2
1 L KCl
0 .150 mol
 0 .350 L KCl
Check:
since need 2x moles of KCl as Pb(NO3)2, and
the molarity of Pb(NO3)2 > KCl, the volume of
KCl should be more than 2x volume Pb(NO3)2
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
38
Table Salt Dissolving in Water
Tro, Chemistry: A Molecular Approach
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
The result is a solution
with free moving
charged particles able
to conduct electricity 39
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
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40
Molecular View of
Electrolytes and Nonelectrolytes
• in order to conduct electricity, a material must have
•
charged particles that are able to flow
electrolyte solutions all contain ions dissolved in the
water
 ionic compounds are electrolytes because they all dissociate
into their ions when they dissolve
• nonelectrolyte solutions contain whole molecules
dissolved in the water
 generally, molecular compounds do not ionize when they
dissolve in water
 the notable exception being molecular acids
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41
Salt vs. Sugar Dissolved in Water
ionic compounds dissociate
into ions when they dissolve
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molecular compounds do not
dissociate when they dissolve
42
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
• 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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43
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
Na2SO4(aq)  2 Na+(aq) + SO42-(aq)
HC2H3O2(aq)  H+(aq) + C2H3O2-(aq)
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Classes of Dissolved Materials
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45
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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46
When Will a Salt Dissolve?
• Predicting whether a compound will dissolve in
water is not easy
• The best way to do it is to do some experiments
to test whether a compound will dissolve in
water, then develop some rules based on those
experimental results
we call this method the empirical method
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47
Solubility Rules
Compounds that Are Generally Soluble in Water
Compounds Containing the Exceptions
Following Ions are Generally (when combined with ions on the
Soluble
left the compound is insoluble)
Li+, Na+, K+, NH4+
none
NO3–, C2H3O2–
none
Cl–, Br–, I–
Ag+, Hg22+, Pb2+
SO42–
Ag+, Ca2+, Sr2+, Ba2+, Pb2+
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48
Solubility Rules
Compounds that Are Generally Insoluble
Exceptions
Compounds Containing the (when combined with ions on the
Following Ions are Generally left the compound is soluble or
Insoluble
slightly soluble)
OH–
S2–
CO32–, PO43–
Tro, Chemistry: A Molecular Approach
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+,
Ca2+, Sr2+, Ba2+
Li+, Na+, K+, NH4+
49
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
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50
2 KI(aq) + Pb(NO3)2(aq)  PbI2(s) + 2
KNO3(aq)
51
No Precipitate Formation =
No Reaction
KI(aq) + NaCl(aq)  KCl(aq) + NaI(aq)
all ions still present,  no reaction
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Process for Predicting the Products of
a Precipitation Reaction
1. Determine what ions each aqueous reactant has
2. Determine formulas of possible products
 Exchange ions
 (+) ion from one reactant with (-) ion from other
 Balance charges of combined ions to get formula of each
product
3. Determine Solubility of Each Product in Water
 Use the solubility rules
 If product is insoluble or slightly soluble, it will precipitate
4. If neither product will precipitate, write no reaction
after the arrow
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53
Process for Predicting the Products of
a Precipitation Reaction
5. If either product is insoluble, write the formulas
for the products after the arrow – writing (s)
after the product that is insoluble and will
precipitate, and (aq) after products that are
soluble and will not precipitate
6. Balance the equation
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54
Example 4.10 – Write the equation for the
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

cross charges and reduce
K2CO3(aq) + NiCl2(aq)  KCl + NiCO3
Example 4.10 – Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
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
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56
Example 4.10 – Write the equation for the
precipitation reaction between an aqueous solution
of potassium carbonate and an aqueous solution of
nickel(II) chloride
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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Ionic Equations
• equations which 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 which describe the actual dissolved species are
called complete ionic equations
 aqueous strong electrolytes are written as ions
 soluble salts, strong acids, strong bases
 insoluble substances, weak electrolytes, and nonelectrolytes
written in molecule form
 solids, liquids, and gases are not dissolved, therefore molecule form
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq)  2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s)
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Ionic Equations
• ions that are both reactants and products are called
spectator ions
2K+1(aq) + 2OH-1(aq) + Mg+2(aq) + 2NO3-1(aq)  2K+1(aq) + 2NO3-1(aq) + Mg(OH)2(s)
• an ionic equation in which the spectator ions are
removed is called a net ionic equation
2OH-1(aq) + Mg+2(aq)  Mg(OH)2(s)
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Acid-Base Reactions
• 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)
• 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
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Acids and Bases in Solution
• acids ionize in water to form H+ ions
 more precisely, the H from the acid molecule is donated to a
water molecule to form hydronium ion, H3O+
 most chemists use H+ and H3O+ interchangeably
• bases dissociate in water to form OH ions
 bases, like NH3, that do not contain OH ions, produce OH by
pulling H off water molecules
• 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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Common Acids
Chemical Name
Formula
Uses
Strength
Perchloric Acid
HClO4
explosives, catalyst
Strong
Nitric Acid
HNO3
explosives, fertilizer, dye, glue
Strong
Sulfuric Acid
H2SO4
Hydrochloric Acid
HCl
explosives, fertilizer, dye, glue,
batteries
metal cleaning, food prep, ore
refining, stomach acid
fertilizer, plastics & rubber,
Strong
Strong
Phosphoric Acid
H3PO4
Chloric Acid
HClO3
Acetic Acid
HC2H3O2
Hydrofluoric Acid
HF
metal cleaning, glass etching
Weak
Carbonic Acid
H2CO3
soda water
Weak
Hypochlorous Acid
HClO
sanitizer
Weak
Boric Acid
H3BO3
eye wash
Weak
food preservation
explosives
plastics & rubber, food
preservation, vinegar
Moderate
Moderate
Weak
Common Bases
Chemical
Name
sodium
hydroxide
potassium
hydroxide
calcium
hydroxide
sodium
bicarbonate
magnesium
hydroxide
Formula
NaOH
Common
Name
Uses
lye,
soap, plastic,
caustic soda
petrol refining
soap, cotton,
Strength
Strong
KOH
caustic potash
Ca(OH)2
slaked lime
cement
Strong
NaHCO3
baking soda
cooking, antacid
Weak
antacid
Weak
Mg(OH)2
milk of
magnesia
ammonium
NH4OH,
ammonia
hydroxide
{NH3(aq)}
water
Tro, Chemistry: A Molecular Approach
electroplating
Strong
detergent,
fertilizer,
Weak
explosives, fibers
63
HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)
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64
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
2.
HNO3(aq) + Ca(OH)2(aq) 
Determine the possible products
a) Determine the ions present when each reactant dissociates
(H+ + NO3-) + (Ca+2 + OH-) 
b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)
(H+ + NO3-) + (Ca+2 + OH-)  (Ca+2 + NO3-) + H2O(l)
c) Write the formula of the salt
 cross the charges
(H+ + NO3-) + (Ca+2 + OH-)  Ca(NO3)2 + H2O(l)
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65
Example - Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide
3. Determine the solubility of the salt
Ca(NO3)2 is soluble
4. Write an (s) after the insoluble products and a
(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)
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66
Example - Write the molecular, ionic, and netionic equation for the reaction of aqueous nitric
acid with aqueous calcium hydroxide
6. Dissociate all aqueous strong electrolytes to
get complete ionic equation
 not H2O
2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq) 
Ca+2(aq) + 2 NO3-(aq) + H2O(l)
7. Eliminate spectator ions to get net-ionic
equation
2 H+1(aq) + 2 OH-1(aq)  2 H2O(l)
H+1(aq) + OH-1(aq)  H2O(l)
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Titration
• often in the lab, a solution’s concentration is
•
determined by reacting it with another material
and using stoichiometry – this process is called
titration
in the titration, the unknown solution is added
to a known amount of another reactant until
the reaction is just completed, at this point,
called the endpoint, the reactants are in their
stoichiometric ratio
 the unknown solution is added slowly from an
instrument called a burette
 a long glass tube with precise volume markings that
allows small additions of solution
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Acid-Base Titrations
• the difficulty is determining when there has been just
enough titrant added to complete the reaction
 the titrant is the solution in the burette
• in acid-base titrations, because both the reactant and
product solutions are colorless, a chemical is added that
changes color when the solution undergoes large
changes in acidity/alkalinity
 the chemical is called an indicator
• at the endpoint of an acid-base titration, the number of
moles of H+ equals the number of moles of OH
 aka the equivalence point
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69
Titration
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70
Titration
The base solution is the
titrant in the burette.
As the base is added to
the acid, the H+ reacts with
the OH– to form water.
But there is still excess
acid present so the color
does not change.
At the titration’s endpoint,
just enough base has been
added to neutralize all the
acid. At this point the
indicator changes color.
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71
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
• Write down the given quantity and its units.
Given:
10.00 mL HCl
12.54 mL of 0.100 M NaOH
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72
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
• Write down the quantity to find, and/or its units.
Find: concentration HCl, M
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73
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
• Collect Needed Equations and Conversion Factors:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
 1 mole HCl = 1 mole NaOH
0.100 M NaOH 0.100 mol NaOH  1 L sol’n
Molarity 
moles solute
liters solution
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74
Example 4.14:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
• Write a Concept Plan:
mL
NaOH
mL
HCl
L
NaOH
mol
NaOH
mol
HCl
0.001 L
0.100 mol NaOH
1 mol HCl
1 mL
1 L NaOH
1 mol NaOH
0.001 L
1 mL
Tro, Chemistry: A Molecular Approach
L
HCl
Molarity 
moles HCl
liters HCl
75
Example:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL of 0.100 M NaOH
Find: M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Apply the Solution Map:
12.54 mL NaOH 
0.001 L
1 mL

0.100 mol NaOH
1L

1 mol HCl
1 mole NaOH
= 1.25 x 10-3 mol HCl
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76
Example:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Apply the Concept Plan:
10.00 mL NaOH 
0.001 L
 0.01000 L HCl
1 mL
Molarity 
1.25 x 10
-3
moles HCl
 0.125 M
0.01000 L HCl
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77
Example:
The titration of 10.00 mL of HCl
solution of unknown concentration
requires 12.54 mL of 0.100 M
NaOH solution to reach the end
point. What is the concentration of
the unknown HCl solution?
Information
Given: 10.00 mL HCl
12.54 mL NaOH
Find: M HCl
CF:
1 mol HCl = 1 mol NaOH
0.100 mol NaOH = 1 L
M = mol/L
CP: mL NaOH → L NaOH →
mol NaOH → mol HCl;
mL HCl → L HCl & mol  M
• Check the Solution:
HCl solution = 0.125 M
The units of the answer, M, are correct.
The magnitude of the answer makes sense since
the neutralization takes less HCl solution than
NaOH solution, so the HCl should be more concentrated.
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Gas Evolving Reactions
• Some reactions form a gas directly from the ion
exchange
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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NaHCO3(aq) + HCl(aq)  NaCl(aq) + CO2(g) + H2O(l)
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Compounds that Undergo
Gas Evolving Reactions
Reactant
Type
Reacting
With
Ion
Exchange
Product
Decompose?
Gas
Formed
Example
metalnS,
metal HS
acid
H2S
no
H2S
K2S(aq) + 2HCl(aq) 
2KCl(aq) + H2S(g)
metalnCO3,
metal HCO3
acid
H2CO3
yes
CO2
K2CO3(aq) + 2HCl(aq) 
2KCl(aq) + CO2(g) + H2O(l)
metalnSO3
metal HSO3
acid
H2SO3
yes
SO2
K2SO3(aq) + 2HCl(aq) 
2KCl(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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81
Example 4.15 - 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
(Na+1 + CO3-2) + (H+1 + NO3-1) 
b) Exchange the anions
(Na+1 + CO3-2) + (H+1 + NO3-1)  (Na+1 + NO3-1) + (H+1 + CO3-2)
c) Write the formula of compounds
 cross the charges
Na2CO3(aq) + HNO3(aq)  NaNO3 + H2CO3
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Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
3. Check to see either product H2S - No
4. Check to see if either product decomposes –
Yes
 H2CO3 decomposes into CO2(g) + H2O(l)
Na2CO3(aq) + HNO3(aq)  NaNO3 + CO2(g) + H2O(l)
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Example 4.15 - When an aqueous solution of
sodium carbonate is added to an aqueous solution
of nitric acid, a gas evolves
5. Determine the solubility of other product
NaNO3 is soluble
6. Write an (s) after the insoluble products and a
(aq) after the soluble products
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3(aq) + CO2(g) + H2O(l)
7. Balance the equation
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO3 + CO2(g) + H2O(l)
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Other Patterns in Reactions
• the precipitation, acid-base, and gas evolving
reactions all involved 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)
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Combustion as Redox
2 H2(g) + O2(g)  2 H2O(g)
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Redox without Combustion
2 Na(s) + Cl2(g)  2 NaCl(s)
2 Na  2 Na+ + 2 e
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Cl2 + 2 e  2 Cl
87
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 reaction involves a metal reacting with a nonmetal
in addition, both reactions involve the conversion of
free elements into ions
4 Na(s) + O2(g) → 2 Na+2O– (s)
2 Na(s) + Cl2(g) → 2 Na+Cl–(s)
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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
89
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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90
Rules for Assigning Oxidation States
• rules are in order of priority
1. free elements have an oxidation state = 0
 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
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Rules for Assigning Oxidation States
3. (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
4. (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
Group 5A
-3
NH3
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Practice – Assign an Oxidation State to
Each Element in the following
•
•
•
•
•
•
Br2
K+
LiF
CO2
SO42-
Na2O2
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94
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)
Na2O2
Na = +1, (Rule 4a) & O = -1, (Rule 3a)
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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 + 2 H2O
-4 +1
0
+4 –2
+1 -2
oxidation
reduction
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96
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
 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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97
Identify the Oxidizing and Reducing Agents
in Each of the Following
3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
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98
Identify the Oxidizing and Reducing Agents
in Each of the Following
red ag
ox ag
+1 -2
+5 -2
3 H2S + 2 NO3– + 2 H+  3 S + 2 NO + 4 H2O
+1
0
+2 -2
+1 -2
oxidation
reduction
ox ag
red ag
+4 -2
+1 -1
MnO2 + 4 HBr  MnBr2 + Br2 + 2 H2O
+2 -1
0
+1 -2
oxidation
reduction
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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 oxidationreduction reactions
2 C8H18(g) + 25 O2(g)  16 CO2(g) + 18 H2O(g)
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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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Practice – Complete the Reactions
• combustion of C3H7OH(l)
• combustion of CH3NH2(g)
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102
Practice – Complete the Reactions
C3H7OH(l) + 5 O2(g)  3 CO2(g) + 4 H2O(g)
CH3NH2(g) + 3 O2(g)  CO2(g) + 2 H2O(g) + NO2(g)
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