Homework Problems

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Homework Problems Chapter 3 Homework Problems
2, 6 (a,b), 10, 24, 31, 34, 40, 44 (a,b), 50, 52,
54, 56, 64, 70, 76, 80, 84, 88, 92, 94, 100, 106,
116, 128
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CHAPTER 3 Stoichiometry Ratios of Combination
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Molecular Mass and Formula Mass The molecular
mass (M) of a substance is the average mass of
one molecule of the substance. For substances
that do not exist as molecules, such as ionic
compounds, we use the formula mass, which
represents the average mass of one formula unit
of the compound. Example Find the formula
mass for potassium chloride (KCl). KCl K 1 .
39.0983 amu 39.0983 amu Cl 1 . 35.453
amu 35.453 amu M 74.551 amu
? 74.55 amu We often round off
molecular mass or formula mass to two places to
the right of the decimal point.
4
Percent By Mass The percent by mass represents
the percentage of each element present in a
sample of a pure chemical substance. The percent
by mass of a particular element is simply the
mass of that element found in one molecule
(formula unit) divided by the total mass of the
molecule (or formula unit), converted to a
percentage by multiplying by 100. percent X by
mass mass of X 100
formula mass Note that the sum of the percent
by mass of each element present in a substance
should be equal to 100 (to within roundoff
error). Example What is the molecular mass, and
the percent by mass of carbon, hydrogen, and
oxygen, in propionaldehyde (C3H6O)?
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Example What is the molecular mass, and the
percent by mass of carbon, hydrogen, and oxygen,
in propionaldehyde (C3H6O)? First step Find
the average mass of each element per molecule and
the average mass per molecule (or formula
unit). C3H6O C 3 . 12.0107 amu 36.0321
amu H 6 . 1.00794 amu 6.04764 amu O 1 .
15.9994 amu 15.9994 amu M
58.0791 amu ? 58.08 amu
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Second step Find the percent composition.
C3H6O C 3 . 12.0107 amu 36.0321 amu H 6
. 1.00794 amu 6.04764 amu O 1 . 15.9994
amu 15.9994 amu M 58.0791
amu ? 58.08 amu C by mass 36.0321
amu 100 62.04 C by mass
58.0791 amu H by mass 6.04764 amu 100
10.41 H by mass 58.0791 amu O by
mass 15.9994 amu 100 27.55 O by
mass 58.0791 amu Note 62.04 10.41
27.55 100.00, as expected.
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Use of Percent Composition as a Conversion
Factor The percent composition of a chemical
substance can be used as a conversion factor in
calculations. Example Titanium metal is often
extracted from rutile, an ore that is 59.9
titanium by mass. What mass of rutile would be
needed to produce 100.0 kg of titanium metal?
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Example Titanium metal is often extracted from
rutile, an ore that is 59.9 titanium by mass.
What mass of rutile would be needed to produce
100.0 kg of titanium metal? The percent
composition tells us that for every 100.0 kg of
rutile we will have 59.9 kg of titanium.
Therefore kg rutile 100.0 kg Ti 100.0 kg
rutile 167. kg rutile 59.9 kg Ti
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Balanced Chemical Equation A balanced chemical
equation indicates the starting substances
(reactants) and ending substances (products) of a
chemical reaction. The reactions may also
indicate the state of the reactants and products,
and/or the physical process used in the
reaction. s solid ? liquid g gas aq
aqueous ? heat A balanced equation must
satisfy conservation of mass (same number of
atoms of each element on the reactant and product
side) and conservation of electrical charge (same
total charge on the reactant and product side).
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Examples of Balanced Chemical Equation Combustion
of hydrogen (combustion is the reaction of a
substance with molecular oxygen to form
combustion products). 2 H2(g) O2(g) ? 2
H2O(?) Nitration of toluene to form
trinitrotoluene (TNT) C6H5CH3 3 HNO3 ?
C6H2CH3(NO2)3 3 H2O Decomposition of calcium
carbonate (decomposition is the breakdown of a
single reactant into two or more
products). CaCO3(s) ? CaO(s)
CO2(g) Dissolution of potassium sulfate, an
ionic compound, in water. K2SO4(s) H2O 2
K(aq) SO42-(aq)
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Procedure For Balancing Chemical Equations 1)
Begin with a complete set of reactants and
products. These cannot be changed. 2) Find a
set of numbers (stoichiometric coefficients) that
satisfy mass balance. a) Change the
coefficients of compounds before changing the
coefficients of elements (save elements for last
in balancing). b) Treat polyatomic ions that
appear on both sides of the equation as a
unit. c) Count atoms and polyatomic ions
carefully, and track them every time you change a
coefficient.
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Procedure For Balancing Chemical Equations 3)
Convert the stoichiometric coefficients into the
smallest set of whole numbers that correctly
balance the chemical equation. 1/2 N2 3/2 H2
? NH3 incorrect fractional coefficient N2 3
H2 ? 2 NH3 correct 2 N2 6 H2 ? 4
NH3 incorrect not smallest set of whole
numbers This step can be saved for last.
Remember that if a chemical equation is correctly
balanced then multiplying all of the coefficients
by the same number results in an equation that is
also balanced. Later in the semester we will
come across a few cases where we do sometimes use
fractional coefficients to balance chemical
equations.
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Example Balance the following unbalanced
chemical equation for the com-bustion of
iso-octane, a hydrocarbon used to determine the
octane rating of gasoline C8H18 O2 ? CO2
H2O (unbalanced)
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Combustion of Iso-octane unbalanced C8H18
O2 ? CO2 H2O balance for C C8H18
O2 ? 8 CO2 H2O balance for H C8H18
O2 ? 8 CO2 9 H2O balance for O C8H18
25/2 O2 ? 8 CO2 9 H2O convert to whole
numbers (multiply through by 2) 2 C8H18
25 O2 ? 16 CO2 18 H2O
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Example - Precipitation of Lead (II)
Chloride Precipitation reaction - A reaction
where two soluble compounds are mixed and a solid
product is formed. Pb(NO3)2(aq)
NaCl(aq) ? PbCl2(s) NaNO3(aq) (unbalanced)
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Example - Precipitation of Lead (II)
Chloride unbalanced Pb(NO3)2(aq) NaCl(aq) ?
PbCl2(s) NaNO3(aq) balance for Cl
Pb(NO3)2(aq) 2 NaCl(aq) ? PbCl2(s)
NaNO3(aq) Balancing for Cl has now caused the
above reaction to be unbal-anced for Na. So we
go back and rebalance for Na
Pb(NO3)2(aq) 2 NaCl(aq) ? PbCl2(s) 2
NaNO3(aq) The above is now balanced. Remember
that when you have a cation or anion group (such
as NO3- above) it is often easier to balance
equations treating this group as a unit.
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The Mole It is nearly impossible to work with
individual atoms in the laboratory because of
their small size and mass. For example, one 12C
atom has a mass of 1.993 x 10-26 kg, far to small
to measure directly.
It is convenient to have a unit representing a
number of atoms that can easily be measured by
the usual techniques in the laboratory. This
unit is called the mole.
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Avogardos Number The unit mole is simply a
number. By definition, one mole of anything is
equal to Avogadros number of things. Avogadros
number, an experimentally determined quantity is
NA 6.022 x 1023. Note that there is no
difference between the concept of moles and more
common terms used for specific numbers of
things. 1 dozen 12 1 score 20 1 thousand
1000 1 mole 6.022 x 1023
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Example (Avogadros Number) If I have three
dozen eggs, then how many eggs do I have? 1
dozen eggs 12 eggs (a conversion
factor) number of eggs 3 dozen eggs 12 eggs
36 eggs
1 dozen If I have three moles of
carbon atoms, then how many carbon atoms do I
have 1 mole carbon atoms 6.022 x 1023 carbon
atoms) number of carbon atoms 3 moles C 6.022
x 1023 C atoms
1
mole 1.81 x 1024 carbon atoms Moles is
the fundamental SI unit for quantity of
substance. The symbol for moles is n, and the
abbreviation for the unit moles is mol.
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Significance of Avogadros Number There must be
something special about the number 6.022 x 1023
(Avogadros number). The significance is as
follows. Consider a collection of identical
objects. The following relationship will
apply. If one object has a mass of X
amu then one mole of objects has a mass of X
g. This is a subtle point. Consider the unit
thousand. We could make the following
statement If one object has a mass of X
g then one thousand objects has a mass of X
kg. Example The mass of one penny is 2.50 g.
The mass of one thousand pennies is mass one
thousand pennies 1000 (2.50 g) 2500. g 1 kg
2.50 kg
1000 g
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The same principle applies to the concept of
moles mass of one 19F atom 19.00 amu,
so mass of one mole of 19F atoms 19.00
g. Check mass one mole 19F (6.022 x 1023)
(19.00 amu) 1.1442 x 1025 amu
1.6605 x 10-27 kg

1 amu 0.01900 kg 19.00 g
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Use of the Mole Concept 1) We may use the
concept of moles to reinterpret atomic mass.
Atomic mass of S 32.06 amu 32.06 g/mol We
interpret the atomic mass of sulfur (or any other
element in the periodic table) in two equivalent
ways The average mass of one sulfur atom is
32.06 amu. The mass of one mole of sulfur atoms
is 32.06 g.
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Use of the Mole Concept 2) We may use the
concept of moles to reinterpret a balanced
chemical equation CH4(g) 2 O2(g) ?
CO2(g) 2 H2O(?) We interpret the balanced
chemical equation in two equivalent ways One
molecule of methane reacts with two molecules of
molecular oxygen to form one molecule of carbon
dioxide and two molecules of water. One mole of
methane reacts with two moles of molecular oxygen
to form one mole of carbon dioxide and two moles
of water. Note that when the mole concept is
used fractional coefficients make sense.
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Use of the Mole Concept (Example) A chemist has
a 14.38 g sample of carbon tetrachloride (CCl4).
a) How many moles of CCl4 does she have? b)
How many atoms of C and Cl are present in the
sample?
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Use of the Mole Concept (Example) A chemist has
a 14.38 g sample of carbon tetrachloride (CCl4).
a) How many moles of CCl4 does she have? CCl4
C 1 . 12.0107 amu 12.0107 amu Cl 4
. 35.453 amu 141.81 amu M
153.82 amu 153.82 g/mol Moles
CCl4 14.38 g 1 mol 0.09349 mol CCl4
153.82 g
9.349 x 10-2 mol CCl4
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Use of the Mole Concept (Example) b) How many
atoms of C and Cl are present in the sample?
C atoms 9.349 x 10-2 mol CCl4 1 mol C
6.022 x 1023 atoms C
1 mol CCl4
mol 5.630 x 1022 C atoms Cl atoms
9.349 x 10-2 mol CCl4 4 mol Cl 6.022 x
1023 atoms C
1 mol CCl4 mol
2.252 x 1023 Cl atoms
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Experimental Determination of the Empirical
Formula The empirical formula for a pure
substance can be found if we know either the
percent composition of the substance (percent by
mass of each element making up the substance) or
the mass of each element present in a sample of
the substance of known mass. The general
procedure we use is as follows 1) Determine the
number of moles of each element present in a
sample of the substance (if you know the percent
by mass of each element it is convenient to
assume 100.0 g of the substance). 2) Divide the
results by whichever number of moles is smallest.
This will give you the number of moles of each
element relative to this smallest number of
moles. 3) Convert the relative number of moles
into the smallest set of whole number
coefficients. This may require multiplication by
a small whole number (2, 3, ).
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Example A 3.862 g sample of a pure chemical
substance was analyzed and found to contain three
elements in the following amounts. Cobalt
(Co) 1.245 g Nitrogen (N) 0.587 g Oxygen
(O) 2.030 g What is the empirical formula of
the substance?
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Cobalt (Co) 1.245 g Nitrogen (N) 0.587
g Oxygen (O) 2.030 g 1) Find the number of
moles of each element. moles Co 1.245 g Co
1 mol 0.02113 mol Co
58.93 g moles N
0.587 g N 1 mol 0.0419 mol N
14.01
g moles O 2.030 g O 1 mol 0.1269 mol
O
16.00 g
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2) Divide through by the smallest number of moles
(0.02113) rel. moles Co 0.02113 1.000
0.02113 rel.
moles N 0.0419 1.98
0.02113 rel. moles O
0.1269 6.006
0.02113 3) Since these numbers are close to
being integer values, the empirical formula
is CoN2O6 (actually Co(NO3)2, cobalt (II)
nitrate)
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Non-Integer Relative Amounts If the relative
amounts of each element present are not close to
integers, try multiplying all of the relative
amounts by a small whole number (2, 3, or 4). If
this gives integer values for the relative
amounts of each element present, these values
represent the empirical formula. Example A
compound containing chromium, oxygen, and
potassium was analyzed, and the following
percentages by mass for each element were found.
Cr 35.35 O 37.99 K 26.66 What is the
empirical formula for the compound?
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Example A compound containing chromium, oxygen,
and potassium was analyzed, and the following
percentages by mass for each element were found.
Cr 35.35 O 37.99 K 26.66 What is the
empirical formula for the compound? Assume 100.0
g of compound. Then moles Cr 35.35 g Cr 1
mol 0.6798 mol Cr
52.00 g moles O
37.99 g O 1 mol 2.374 mol O
16.00
g moles K 26.66 g K 1 mol 0.6818 mol
K
39.10 g
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Divide through by the smallest number of moles
to get the relative number of moles of each
element present rel moles Cr 0.6798 mol
1.000 0.6798
mol rel moles O 2.374 mol 3.492
0.6798 mol rel moles
K 0.6818 mol 1.003
0.6798 mol Since these are not close
to integer values, try multiplying by 2 rel mol
Cr 2.000 rel mol O 6.984 rel mol K
2.006 So the empirical formula is Cr2O7K2
(actually K2Cr2O7, potassium dichromate)
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Combustion Analysis Combustion analysis is a
method often used in the analysis of organic
compounds. A sample of the compound is allowed
to completely react with oxygen and the mass of
carbon dioxide (CO2) and water (H2O) produced is
measured. From this information the number of
grams and the number of moles of carbon and
hydrogen originally in the sample can be found.
For hydrocarbons (compounds containing only
hydrogen and carbon) this is sufficient to find
the empirical formula. For molecules containing
other elements additional analysis is required.
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Example A pure hydrocarbon is analyzed by
combustion analysis. Combustion of a sample of
the hydrocarbon produces 3.65 g CO2 and 1.99 g
H2O. What is the empirical formula for the
hydrocarbon? Note M(CO2) 44.01 g/mol M(H2O)
18.02 g/mol.
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Example A pure hydrocarbon is analyzed by
combustion analysis. Combustion of a sample of
the hydrocarbon produces 3.65 g CO2 and 1.99 g
H2O. What is the empirical formula for the
hydrocarbon? Note M(CO2) 44.01 g/mol M(H2O)
18.02 g/mol. moles C 3.65 g CO2 1 mol
CO2 1 mol C 0.0829 mol C
44.01 g CO2 1 mol CO2
moles H 1.99 g H2O 1 mol H2O 2 mol
H 0.221 mol H
18.02 g H2O 1 mol H2O relative moles
C 0.0829 mol/0.0829 mol 1.00 (3.00) relative
moles H 0.221 mol/0.0829 mol 2.67 (8.00) To
get values close to integers multiply by
3. Empirical formula C3H8
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Finding the Molecular Formula From the Empirical
Formula As previously discussed, there is a
relationship between the empirical formula and
the molecular formula for the substance. (molecul
ar formula) N (empirical formula) where N is an
integer (1, 2, 3, ) Substance Molecular Empiric
al N formula formula propane C4H10 C2H5
2 sulfuric acid H2SO4 H2SO4 1
cyclohexane C6H12 CH2 6
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Finding the Molecular Formula From the Empirical
Formula Since (molecular formula) N
(empirical formula) Then N mass of molecular
formula mass of empirical formula So
if we can find the empirical formula and the mass
of the molecule we can find N, and from that the
molecular formula. Example The empirical
formula for a substance is CH2O. The molecular
mass of the substance is 60. g/mol. What is the
molecular formula for the substance?
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Finding the Molecular Formula From the Empirical
Formula Example The empirical formula for a
substance is CH2O. The molecular mass of the
substance is 60. g/mol. What is the molecular
formula for the substance? N mass of molecular
formula mass of empirical formula The
mass of the empirical formula is
M(empirical) 1 (12.01 g/mol) 2 (1.008 g/mol)
1 (16.00 g/mol) 30.03 g/mol N 60.
g/mol 2.0 30.03 g/mol So the molecular
formula is C2H4O2.
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Interpretation of Chemical Equations Consider
the equation for the formation of water from
hydrogen and oxygen, discussed previously. 2
H2(g) O2(g) ? 2 H2O(?) We can interpret
this equation in two ways 1) Two molecules of
hydrogen and one molecule of oxygen combine to
form two molecules of water. 2) Two moles of
hydrogen and one mole of oxygen cobbine to form
two moles of water. Note that the second
interpretation makes sense even if we have
fractional stoichiometric coefficients.
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One of the most important uses of a balanced
chemical equation is to provide the information
needed to convert from the amount of one
substance inolved in a reaction and the amount of
a second substance that is required for the
reaction. Example Consider the following
balanced chemical equation 2 H2(g) O2(g) ?
2 H2O(?) If we begin with 4.8 moles of H2, how
many moles of O2 will be required for a complete
reaction? What is the maximum number of moles of
H2O that we will be able to produce?
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Example Consider the following balanced
chemical equation 2 H2(g) O2(g) ? 2
H2O(?) If we begin with 4.8 moles of H2, how
many moles of O2 will be required for a complete
reaction? What is the maximum number of moles of
H2O that we will be able to produce? mol O2
4.8 mol H2 1 mol O2 2.4 mol O2
2 mol H2 mol
H2O 4.8 mol H2 2 mol H2O 4.8 mol H2O
2 mol
H2
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Example When sodium metal is placed in water
a violent reaction occurs. 2 Na(s) 2 H2O(?)
? 2 NaOH(aq) H2(g) If we begin with 4.0
moles of sodium metal, how many moles of water
are required to completely react with the sodium
metal? How many moles of hydrogen gas will form?
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2 Na(s) 2 H2O(?) ? 2 NaOH(aq) H2(g) If
we begin with 4.0 moles of sodium metal, how many
moles of water are required to completely react
with the sodium metal? How many moles of
hydrogen gas will form? mol H2O 4.0 mol Na 2
mol H2O 4.0 mol H2O
2 mol Na mol H2
4.0 mol Na 1 mol H2 2.0 mol H2
2 mol Na
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Calculations Using Mass of Reactants and
Products In the laboratory we work with masses
of reactants and products. However, as we have
seen, chemical equations give relationships among
moles of reactants and products. We therefore
need a method for going back and forth between
moles of substance and grams of substance. We
may do this conversion using M, the molecular
mass or formula mass of a substance. For
example, the molecular mass of ethane (C2H6)
is M(C2H6) 30.07 g/mol This is a conversion
factor between mass of C2H6 and moles of C2H6.
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Strategy For Calculations Using Balanced Chemical
Equations We may use the following strategy to
find the mass of reactant or product in a
chemical equation. 1) Convert from grams of the
first substance into moles of the first substance
using the molecular mass of the substance. 2)
Use the balanced chemical equation to find the
correct number of moles of the second
substance. 3) Convert from number of moles of
the second substance to mass of the second
substance using the molecular mass of the
substance.
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Example We can use a balanced chemical equation
and our new concept of moles to do calculations
involving chemical reactions. Example Calcium
chloride (CaCl2) can be produced by the reaction
of hydrochloric acid (HCl) and calcium hydroxide
(Ca(OH)2). 2 HCl Ca(OH)2 ? CaCl2 2
H2O In a particular reaction we begin with
10.00 g HCl. How many grams of Ca(OH)2 are
required to completely react with this amount of
HCl? How many grams of calcium chloride can be
produced by the reaction?
calcium chloride
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2 HCl Ca(OH)2 ? CaCl2 2 H2O In a
particular reaction we begin with 10.00 g HCl.
How many grams of Ca(OH)2 are required to
completely react with this amount of HCl? How
many grams of calcium chloride can be produced by
the reaction? M(HCl) 36.46 g/mol M(CaCl2)
110.98 g/mol M(Ca(OH)2) 74.09 g/mol M(H2O)
18.02 g/mol 1) Find the moles of HCl. n(HCl)
10.00 g HCl 1 mol HCl 0.2743 mol
HCl
36.46 g HCl 2) Find the moles Ca(OH)2.
n(Ca(OH)2) 0.2743 mol HCl 1 mol Ca(OH)2
0.1371 mol Ca(OH)2
2 mol HCl
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3) Find the grams of Ca(OH)2 m(Ca(OH)2)
0.1371 mol Ca(OH)2 74.09 g Ca(OH)2 10.16 g

1 mol Ca(OH)2 We can also
find the mass of calcium chloride that can be
produced by the reaction m(CaCl2) 0.2743 mol
HCl 1 mol CaCl2 110.98 g CaCl2
2 mol HCl
1 mol CaCl2 15.22 g CaCl2
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Yields For a Chemical Reaction There are three
types of yields we can define for a chemical
reaction. Theoretical yield - The maximum amount
of product that can be formed from the starting
materials used in the reaction. Actual yield -
The observed yield for a chemical
reaction. Percent yield - The percent of the
theoretical yield that is actually obtained.
yield actual yield . 100
theoretical yield For example, if the amount of
CaCl2 actually obtained in the above reaction was
11.46 g, then the percent yield for the reaction
would be yield 11.46 g . 100 75.3
15.22 g
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Limiting Reactant In most chemical reactions
several reactants combine to form products. As
soon as one of the reactants runs out, the
reaction will stop, even if the other reactants
are still present. We define the limiting
reactant as the reactant the first runs out in a
chemical reaction. Note that the theoretical
yield of product is determined by the limiting
reactant. Example Consider making sandwiches
out of ham and bread. One sandwich consists of
two slides of bread and one slice of ham. If we
begin with 10 slices of bread and 10 slices of
ham, what is the limiting reactant and the
theoretical yield of ham sandwiches?
52
limiting reactant bread theoretical yield
5 sandwiches
53
Example Water molecules are formed from 2
hydrogen atoms and 1 oxygen atom. If we begin
with 10 hydrogen atoms and 10 oxygen atoms, what
is the limiting reactant and the theoretical
yield of water?
limiting reactant hydrogen theoretical yield
5 water molecules Note this works equally well
if done in terms of moles.
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Finding the Limiting Reactant For a reaction
involving two reactants (A and B) the limiting
reactant can be found as follows. 1) Find the
number of moles of A and B. 2) Use the balanced
chemical equation to find the number of moles of
the B required to completely react with the
number of moles of A that are present. There are
three possibilities. a) If the actual number
of moles of B is more than is needed to
completely react with A, then A runs out first
and is the limiting reactant. b) If the actual
number of moles of B is less than is needed to
completely react with A, then B runs out first
and is the limiting reactant. c) If the actual
number of moles of B is exactly what is needed to
completely react with A, then both reactants will
completely react.
55
Example Consider the following chemical
reaction (molecular masses are given below each
reactant). C2H2 2 HCl
? C2H4Cl2 (26.04 g/mol)
(36.46 g/mol) (98.96 g/mol) In a
particular experiment we begin with 10.00 g C2H2
and 20.00 g HCl. What is the limiting reactant
and the theoretical yield of C2H4Cl2?
56
C2H2 2 HCl ?
C2H4Cl2 (26.04 g/mol) (36.46
g/mol) (98.96 g/mol) In a particular
experiment we begin with 10.00 g C2H2 and 20.00 g
HCl. What is the limiting reactant and the
theoretical yield of C2H4Cl2? moles C2H2
10.00 g 1 mol 0.3840 mol C2H2
26.04
g moles HCl 20.00 g 1 mol 0.5485 mol
HCl
36.46 g moles HCl needed for complete
reaction 0.3840 mol C2H2 2 mol HCl
0.7680 mol HCl 1 mol C2H2 Since we
only have 0.5485 mol HCl, HCl is the limiting
reactant.
57
We can now find the theoretical yield of our
product, C2H4Cl2. We have 0.5485 moles of HCl,
our limiting reactant. So theoretical yield
0.5485 mol HCl 1 mol C2H4Cl2 98.96 g
C2H4Cl2
2 mol HCl 1 mol C2H4Cl2
27.14 g C2H4Cl2.
58
Finding the Limiting Reactant For a reaction
involving more than two reactants the limiting
reactant can be found as follows. 1) Find the
number of moles of each reactant. 2) Use the
balanced chemical equation to find the number of
moles of a particular product that can form from
each reactant. 3) Whichever number of moles is
smallest corresponds to the limiting
reactant. Note that this method also works when
there are two reactants.
59
We can use the same example previously used to
show how this method works. C2H2
2 HCl ? C2H4Cl2 (26.04
g/mol) (36.46 g/mol) (98.96
g/mol) In a particular experiment we begin with
10.00 g C2H2 and 20.00 g HCl. What is the
limiting reactant and the theoretical yield of
C2H4Cl2? moles C2H2 10.00 g 1 mol
0.3840 mol C2H2
26.04 g moles HCl 20.00 g
1 mol 0.5485 mol HCl
36.46 g We now find
how many moles of C2H4Cl2 we could form from the
given amounts of each reactant. Whichever number
of moles is smallest will be the theoretical
yield (in moles of product) and the corresponding
reactant will be the limiting reactant.
60
C2H2 2 HCl ?
C2H4Cl2 (26.04 g/mol) (36.46
g/mol) (98.96 g/mol) In a particular
experiment we begin with 10.00 g C2H2 and 20.00 g
HCl. What is the limiting reactant and the
theoretical yield of C2H4Cl2? moles C2H2
0.3840 mol moles HCl 0.5485 mol mol C2H4Cl2
(from C2H2) 0.3840 mol C2H2 1 mol C2H4Cl2

1 mol C2H2
0.3840 mol C2H4Cl2 mol C2H4Cl2 (from
HCl) 0.5485 mol HCl 1 mol C2H4Cl2

2 mol HCl 0.2742
mol C2H4Cl2 Since 0.2742 mol C2H4Cl2 is
smaller, then that represents the theoretical
yield for C2H4Cl2 (in moles), and HCl is the
limiting reactant (as previously found). All we
would do now is convert to grams C2H4Cl2 to get
the theoretical yield in terms of mass of product.
61
Types of Chemical Reactions There are literally
thousands of different types of chemical
reactions. Three common types are Combination
reaction - Two or more reactants form a single
product. C(s) O2(g) ? CO2(g) Decomposition
reaction - A single reactant forms two or more
products. CuSO4 5 H2O(s) ? CuSO4(s) 5
H2O(g) Combustion reaction - Reaction of a
single substance with molecular oxygen to form
combustion products (CO2(g), H2O(?). C3H8(g) 5
O2(g) ? 3 CO2(g) 4 H2O(?)
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End of Chapter 3
I have lived much of my life among molecules.
They are good company. - George Wald The most
exciting phrase to hear in science, the one that
heralds new discoveries, is not Eureka! (I found
it!) but rather, hmm.... that's funny. -
Isaac Asimov it has become hard to find an
important problem (in science) that is not
already being worked on by crowds of people on
several continents. - Max Perutz Not
everyone learns from books. - Sheila
Rowbothum, Resistance and Revolution