Chapter 5 chemical reaction
1
Mole and Avogadro's number
A mole is a quantity that contains 6.02 x 1023 items.
• Just as a grocer sells rice by weight rather than by counting
grains; a chemist uses weight to count for atoms
• As a dozen of anything contains 12 a mole of anything
contains 6.022x1023
Use of the mole?
• 1mole = Avogadro's number
• This graph will help you with most of chapter 5 calculations
The Mole and Avogadro’s Number
It can be used as a conversion factor to relate the
number of moles of a substance to the number of
atoms or molecules:
1 mol
6.02 x 1023 atoms
or
6.02 x 1023 atoms
1 mol
1 mol
or 6.02 x 1023 molecules
6.02 x 1023 molecules
1 mol
4
Mole Calculations in Chemical Equations
Coefficients are used to form mole ratios, which can
serve as conversion factors.
N2(g)
+
O2(g)
2 NO(g)
Mole ratios:
1 mol N2
1 mol O2
1 mol N2
2 mol NO
1 mol O2
2 mol NO
5
Mass Calculations in Chemical Equations
HOW TO Convert Grams of Reactant to Grams of Product
mole–mole
conversion
factor
Moles of
reactant
molar mass
conversion
factor
[1]
Grams of
reactant
[2]
Moles of
product
molar mass
conversion
factor
[3]
Grams of
product
6
5.7 Percent Yield
•The theoretical yield is the amount of product
expected from a given amount of reactant based
on the coefficients in the balanced chemical
equation.
•Usually, however, the amount of product formed
is less than the maximum amount of product
predicted.
•The actual yield is the amount of product isolated
from a reaction.
7
5.7 Percent Yield
Sample Problem 5.14
If the reaction of ethylene with water to form
ethanol has a calculated theoretical yield of 23 g
of ethanol, what is the percent yield if only 15 g
of ethanol are actually formed?
Percent yield
actual yield (g)
= theoretical yield (g) x 100%
=
15 g
23 g x
100% =
65%
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5.8 Limiting Reactants
•The limiting reactant is the reactant that is
completely used up in a reaction.
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Limiting Reactant
A limiting reactant in a chemical reaction is the
substance that
• Is used up first.
• Stops the reaction.
• Limits the amount of product that can form.
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Determining the Limiting Reactant
Sample Problem 5.18
[1]: Determine how much of one reactant is
needed to react with a second reactant.
2 H2(g) + O2(g)
chosen to be
“Original Quantity”
2 H2O(l)
chosen to be
“Unknown Quantity”
There are 4
molecules of H2
in the picture.
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Determining the Limiting Reactant
Sample Problem 5.18
[2]: Write out the conversion factors that relate the
numbers of moles (or molecules) of reactants
2 H2(g) + O2(g)
2 molecules H2
1 molecule O2
2 H2O(l)
1 molecule O2
2 molecules H2
Choose this
conversion factor to
cancel molecules H2
12
Determining the Limiting Reactant
Sample Problem 5.18
[3]: Calculate the number of moles (molecules) of
the second reactant needed for complete reaction.
2 H2(g) + O2(g)
4 molecules H2 x 1 molecule O2
2 molecules H2
2 H2O(l)
= 2 molecules O2
13
Determining the Limiting Reactant
Sample Problem 5.18
[4]: Analyze the two possible outcomes:
• If the amount present of the second reactant
is less than what is needed, the second
reactant is the limiting reagent.
• If the amount present of the second reactant is
greater than what is needed, the second
reactant is in excess.
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Determining the Limiting Reactant
Sample Problem 5.18
15
Determining the Limiting Reactant Using the Number of
Grams
Sample Problem 5.20
Using the balanced equation, determine the limiting
reactant when 10.0 g of N2 (MM = 28.02 g/mol) react
with 10.0 g of O2 (MM = 32.00 g/mol).
N2(g) + O2(g)
2 NO(g)
16
Determining the Limiting Reactant Using the Number of
Grams
Sample Problem 5.20
[1] Convert the number of grams of each reactant
into moles using the molar masses.
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Determining the Limiting Reactant Using the Number of
Grams
Sample Problem 5.20
[2] Determine the limiting reactant by choosing N2
as the original quantity and converting to mol O2.
mole–mole
Conversion factor
0.357 mol N2 x
1 mol O2
1 mol N2
= 0.357 mol O2
The amount of O2 we started with (0.313 mol) is
less than the amount we would need (0.357 mol) so
O2 is the limiting reagent.
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End of chapter 5 question 87
• Question 87
• The local anesthetic ethyl chloride (C2H5Cl, molar mass is
64.51g/mol) can be prepared by reaction of ethylene (C2H4
molar mass 28.05g/mol) according to the balanced equation
C2H4 + HCl
C2H5Cl
• A. if 8.00g of ethylene and 12.0g of HCl are used, how many
moles of each reactant are used?
• What is the limiting reactant
• How many grams of the product formed?
• If a 10.6g of product are formed, what is the percent yield of
the reaction?
Problem 5.87
• The local anesthetic ethyl chloride (C2H5Cl, molar mass
64.51g/mole) can be prepared by reaction of ethylene (C2H4, molar
mass 28.05g/mole) with HCl (molar mass 36.46g/mole), according
to the balanced equation,
C2H4 + HCl
C2H5Cl
• a. if 8.00g of ethylene and 12.0g of HCl are used, how many moles
of each reacted are used?
8.00g C2H4 x
12.0g HCl x
1 mol C2H4
28.05g C2H4
= 0.285 mol C2H4
1 mol HCl
36.46g HCl
= 0.329 mol HCl
Problem 5.87
C2H4 + HCl
C2H5Cl
• b. What is the limiting reactant
• 0.285 mol C2H4 is completely used up in the reaction so it is
the limiting reactant
• c. how many moles of product are formed
0.285mol C2H4 x 1 mol C2H5Cl = 0.285 mol C2H5Cl
1mol C2H4
Problem 5.87
C2H5Cl
C2H4 + HCl
• d. How many grams of the product formed
0.285mol C2H5Cl x 64.51gC2H5Cl
1mol C2H5Cl
= 18.4 g C2H5Cl
• e. if 10.6g of product are formed, what is the percent yield of
the reaction?
actual yield (g)
Percent yield
= theoretical yield (g) x 100%
Percent yield
=
10.6 g C2H5Cl
18.4 g C2H5Cl
Percent yield
=
57.6%
x 100%
Alka seltzer Calculations
• 1. NaHCO3(s)  Na+(aq) + HCO3-(aq)
• 2. HCO3-(aq) + H3O+(aq)  2H2O(l) + CO2 (g)
• For example if you determined the mass of CO2 lost is 0.512g
determine moles of CO2 (g) lost
0.50g CO2 x
1 mol CO2
44.01g CO2
= 0.0163 mol CO2
Use mole ratio to calculate moles of NaHCO3(s)
0.0163moles CO2 x 1 mole NaHCO3(s) = 0.0163 mol NaHCO3(s)
1 mole CO2
Alka seltzer Calculations
• Mass of NaHCO3(s)
0.0163moles NaHCO3 x 83.00g NaHCO3(s) = 1.35g NaHCO3(s)
1 mole NaHCO3(s)
• Calculated Mass% of NaHCO3 reacted in tablet (printed on
label is 1.916g of NaHCO3
1.35g NaHCO3(s)
1.916g in tablet
x 100%
= 70 % NaHCO3(s) reacted
5.9 Oxidation and Reduction
A. General Features
•Oxidation is the loss of electrons from an atom.
•Reduction is the gain of electrons by an atom.
•Both processes occur together in a single reaction
called an oxidation−reduction or redox reaction.
Thus, a redox reaction always has two components,
one that is oxidized and one that is reduced.
•A redox reaction involves the transfer of electrons
from one element to another.
25
5.9 Oxidation and Reduction
Cu2+ gains 2 e−
Zn + Cu2+
Zn2+ + Cu
Zn loses 2 e–
•Zn loses 2 e− to form Zn2+, so Zn is oxidized.
•Cu2+ gains 2 e− to form Cu, so Cu2+ is reduced.
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5.9 Oxidation and Reduction
Cu2+ gains 2 e−
Zn2+ + Cu
Zn + Cu2+
Zn loses 2 e–
Each of these processes can be written as an
individual half reaction:
Zn2+ + 2 e−
loss of e−
Oxidation half reaction:
Zn
Reduction half reaction:
Cu2+ + 2e−
gain of 27
e−
Cu
5.9 Oxidation and Reduction
Zn
+ Cu2+
oxidized reduced
Zn2+ + Cu
A compound that is oxidized while causing another
compound to be reduced is called a reducing agent.
•Zn acts as a reducing agent because it causes
Cu2+ to gain electrons and become reduced.
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5.9 Oxidation and Reduction
Zn
+ Cu2+
oxidized reduced
Zn2+ + Cu
A compound that is reduced while causing another
compound to be oxidized is called an oxidizing agent.
•Cu2+ acts as an oxidizing agent because it causes
Zn to lose electrons and become oxidized.
29
5.9 Oxidation and Reduction
30
Examples of Oxidation–Reduction Reactions
Iron Rusting
O gains e– and is reduced.
4 Fe(s) + 3 O2(g)
neutral Fe neutral O
2 Fe2O3(s)
Fe3+ O2–
Fe loses e– and is oxidized.
31
Examples of Oxidation–Reduction Reactions
Inside an Alkaline Battery
Mn4+ gains e− and is reduced.
Zn + 2 MnO2
neutral Zn
Mn4+
ZnO + Mn2O3
Zn2+
Mn3+
Zn loses e− and is oxidized.
32
Examples of Oxidation–Reduction Reactions
Zn + 2 MnO2
ZnO + Mn2O3
33
Examples of Oxidation–Reduction Reactions
Oxidation results in the:
Reduction results in the:
•Gain of oxygen atoms
•Loss of oxygen atoms
•Loss of hydrogen atoms
•Gain of hydrogen atoms
34
redox chemistry
Oxidation occurs when a
molecule does any of the
following:
•
Loses electrons
• Gains oxygen
If a molecule undergoes oxidation, it is the
reducing agent.
redox chemistry
Reduction occurs when a molecule
does any of the following:
Gains electrons
Loses oxygen
If a molecule undergoes reduction, it is the oxidizing
agent.
Question 5.92
• Identify the species that is oxidized and the species that is
reduced in each reaction. Write out two half reactions to show
how many electrons are gained or lost by each species.
Mg + Fe2+
Mg2+ + Fe
Sn + Cu2+
Sn2+ + Cu
4Na + O2
2Na2O
Mg2+ + Fe
Mg + Fe2+
Fe2+ gains 2 e−
Mg2+ + Fe
Mg + Fe2+
Mg loses 2 e–
Each of these processes can be written as an
individual half reaction:
Mg2+ + 2 e−
loss of e−
Oxidation half reaction:
Mg
Reduction half reaction:
Fe2+ + 2e−
gain of 38
e−
Fe
Sn2+ + Cu
Sn + Cu2+
Cu2+ gains 2 e−
Sn2+ + Cu
Sn + Cu2+
Sn loses 2 e–
Each of these processes can be written as an
individual half reaction:
Sn2+ + 2 e−
loss of e−
Oxidation half reaction:
Sn
Reduction half reaction:
Cu2+ + 2e−
gain of 39
e−
Cu
Chapter 5 question 91
• Identify the species that is oxidized and the species that is
reduced in each reaction. Write out two half reactions to show
how many electrons are gained or lost by each species.
Fe + Cu2+
Cl2 + 2I2Na + Cl2
Fe 2+ + Cu
I2 + 2Cl2NaCl
2Na2O
4Na + O2
Loss of oxygen
2Na2O
4Na + O2
gain of oxygen
Each of these processes can be written as an
individual half reaction:
4Na1+ + 4 e−
loss of e−
Oxidation half reaction:
4Na
Reduction half reaction:
O2 + 4e−
gain of 41
e−
2O-2
Question 5.92
• Identify the species that is oxidized and the species that is reduced
in each reaction. Write out two half reactions to show how many
electrons are gained or lost by each species.
Mg + Fe2+
Mg2+ + Fe
Mg is oxidized becomes the reducing agent
Fe2+ is reduced becomes the oxidizing agent
Sn + Cu2+
Sn2+ + Cu
Zn is oxidized becomes the reducing agent
Cu2+ is reduced becomes the oxidizing agent
4Na + O2
2Na2O
Na is oxidized becomes the reducing agent
O is reduced becomes the oxidizing agent