Transcript Stoichiometry

Chapter Three:
STOICHIOMETRY
講義
Assignment for Chapter 3
13,18,28,52,59,68,78,83,91,101,130
Chapter 3 | Slide 2
Core Materials
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Stoichiometry
Mole
Molar mass
Percent composition
Empirical formula
Molecular formula
Chemical reactions
Characteristics of a chemical
equation
• Stoichiometry calculations
• Yield
Chapter 3 | Slide 3
Chemical Stoichiometry
• Stoichiometry – The study of quantities of
materials consumed and produced in
chemical reactions.
Chapter 3 | Slide 4
Counting by Weighing
• Need average mass of the object.
• Objects behave as though they were all
identical.
3.1
Chapter 3 | Slide 5
Atomic Masses
• Elements occur in nature as mixtures of
isotopes.
• Carbon = 98.89% 12C
1.11% 13C
<0.01% 14C
Carbon atomic mass = 12.01 amu
3.2
Chapter 3 | Slide 6
Schematic Diagram of a Mass
Spectrometer
3.2
Chapter 3 | Slide 7
The Mole
• The number equal to the number of
carbon atoms in exactly 12 grams of pure
12C
• 1 mole of anything = 6.022 x 1023 units of
that thing (Avogadro’s number)
• 1 mole C = 6.022 x 1023 C atoms = 12.01 g C
3.3
Chapter 3 | Slide 8
Molar Mass
• Mass in grams of one mole of the substance:
Molar Mass of N = 14.01 g/mol
Molar Mass of H2O = 18.02 g/mol
(2 × 1.008) + 16.00
Molar Mass of Ba(NO3)2 = 261.35 g/mol
137.33 + (2 × 14.01) + (6 × 16.00)
3.4
Chapter 3 | Slide 9
Concept Check
Which of the following is closest to the average
mass of one atom of copper?
a)
b)
c)
d)
e)
63.55 g
52.00 g
58.93 g
65.38 g
1.055 x 10-22 g
Chapter 3 | Slide 10
Concept Check
Calculate the number of copper atoms in a 63.55 g
sample of copper.
Chapter 3 | Slide 11
Concept Check
Which of the following 100.0 g samples contains
the greatest number of atoms?
a) Magnesium
b) Zinc
c) Silver
Chapter 3 | Slide 12
Exercise
Rank the following according to number of atoms
(greatest to least):
a) 107.9 g of silver
b) 70.0 g of zinc
c) 21.0 g of magnesium
Chapter 3 | Slide 13
Exercise
Consider separate 100.0 gram samples of each of
the following:
H2O, N2O, C3H6O2, CO2
– Rank them from greatest to least number of
oxygen atoms.
Chapter 3 | Slide 14
Percent Composition
• Mass percent of an element:
mass of element in compound
mass % 
 100%
mass of compound
• For iron in iron(III) oxide, (Fe2O3):
111.69
mass % Fe 
 100%  69.94%
159.69
3.5
Chapter 3 | Slide 15
Exercise
Consider separate 100.0 gram samples of each of
the following:
H2O, N2O, C3H6O2, CO2
– Rank them from highest to lowest percent
oxygen by mass.
Chapter 3 | Slide 16
Formulas
• Molecular formula = (empirical formula)n
[n = integer]
• Molecular formula = C6H6 = (CH)6
– Actual formula of the compound
• Empirical formula = CH
– Simplest whole-number ratio
3.6
Chapter 3 | Slide 17
Analyzing for Carbon and Hydrogen
• Device used to determine the mass percent
of each element in a compound.
3.6
Chapter 3 | Slide 18
Balancing Chemical Equations
3.7 and 3.8
Chapter 3 | Slide 19
Chemical Equation
• A representation of a chemical reaction:
C2H5OH + 3O2  2CO2 + 3H2O
reactants
products
• Reactants are only placed on the left side
of the arrow, products are only placed on
the right side of the arrow.
3.7 and 3.8
Chapter 3 | Slide 20
Chemical Equation
C2H5OH + 3O2  2CO2 + 3H2O
• The equation is balanced.
• All atoms present in the reactants are
accounted for in the products.
• 1 mole of ethanol reacts with 3 moles of
oxygen to produce 2 moles of carbon
dioxide and 3 moles of water.
3.7 and 3.8
Chapter 3 | Slide 21
Chemical Equations
• The coefficients in the balanced
equation have nothing to do with the
amount of each reactant that is given in
the problem.
3.7 and 3.8
Chapter 3 | Slide 22
Chemical Equations
• The balanced equation represents a
ratio of reactants and products, not
what actually “happens” during a
reaction.
• Use the coefficients in the balanced
equation to decide the amount of each
reactant that is used, and the amount of
each product that is formed.
3.7 and 3.8
Chapter 3 | Slide 23
Exercise
Which of the following correctly balances the
chemical equation given below? There may be
more than one correct balanced equation. If a
balanced equation is incorrect, explain what is
incorrect about it.
CaO + C  CaC2 + CO2
I.
II.
III.
IV.
Chapter 3 | Slide 24
CaO2 + 3C  CaC2 + CO2
2CaO + 5C  2CaC2 + CO2
CaO + (2.5)C  CaC2 + (0.5)CO2
4CaO + 10C  4CaC2 + 2CO2
Concept Check
Which of the following are true concerning
balanced chemical equations? There may be
more than one true statement.
I. The number of molecules is conserved.
II. The coefficients tell you how much of each
substance you have.
III. Atoms are neither created nor destroyed.
IV. The coefficients indicate the mass ratios of the
substances used.
V.The sum of the coefficients on the reactant side
equals the sum of the coefficients on the product
side.
Chapter 3 | Slide 25
Notice
• The number of atoms of each type of element must be
the same on both sides of a balanced equation.
• Subscripts must not be changed to balance an equation.
• A balanced equation tells us the ratio of the number of
molecules which react and are produced in a chemical
reaction.
• Coefficients can be fractions, although they are usually
given as lowest integer multiples.
3.8
Chapter 3 | Slide 26
Stoichiometric Calculations
• Chemical equations can be used to relate
the masses of reacting chemicals.
3.9
Chapter 3 | Slide 27
Calculating Masses in Reactions
• Balance the equation.
• Convert mass to moles.
• Set up mole ratios from the balanced
equation.
• Calculate number of moles of desired
reactant or product.
• Convert back to grams.
3.9
Chapter 3 | Slide 28
Exercise
• Methane (CH4) reacts with the oxygen in the air
to produce carbon dioxide and water.
• Ammonia (NH3) reacts with the oxygen in the air
to produce nitrogen monoxide and water.
• What mass of ammonia would produce the
same amount of water as 1.00 g of methane
reacting with excess oxygen?
Chapter 3 | Slide 29
Let’s Think About It
We need to know:
– How much water is produced from 1.00 g of
methane and excess oxygen.
– How much ammonia is needed to produce the
amount of water calculated above.
Chapter 3 | Slide 30
Limiting Reactants
• Limiting reactant – the reactant that is
consumed first and therefore limits the
amounts of products that can be formed.
• Determine which reactant is limiting to
calculate correctly the amounts of
products that will be formed.
3.10
Chapter 3 | Slide 31
Limiting Reactants
3.10
Chapter 3 | Slide 32
Limiting Reactants
• Methane and water will react to form
products according to the equation:
CH4 + H2O  3H2 + CO
3.10
Chapter 3 | Slide 33
Mixture of CH4 and H2O Molecules Reacting
Chapter 3 | Slide 34
CH4 and H2O Reacting to Form H2 and CO
Chapter 3 | Slide 35
Limiting Reactants
• The amount of products that can form is
limited by the methane.
• Methane is the limiting reactant.
• Water is in excess.
3.10
Chapter 3 | Slide 36
Concept Check
Which of the following reaction mixtures could
produce the greatest amount of product? Each
involves the reaction symbolized by the equation:
2H2 + O2  2H2O
a)
b)
c)
d)
e)
Chapter 3 | Slide 37
2 moles of H2 and 2 moles of O2
2 moles of H2 and 3 moles of O2
2 moles of H2 and 1 mole of O2
3 moles of H2 and 1 mole of O2
Each produce the same amount of product
Notice
• We cannot simply add the total moles of
all the reactants to decide which reactant
mixture makes the most product. We
must always think about how much
product can be formed by using what we
are given, and the ratio in the balanced
equation.
3.10
Chapter 3 | Slide 38
Concept Check
• You know that chemical A reacts with chemical B.
You react 10.0 g of A with 10.0 g of B.
• What information do you need to know in order
to determine the mass of product that will be
produced?
Chapter 3 | Slide 39
Let’s Think About It
We need to know:
– The mole ratio between A, B, and the product
they form. In other words, we need to know
the balanced reaction equation.
– The molar masses of A, B, and the product
they form.
Chapter 3 | Slide 40
Exercise
• You react 10.0 g of A with 10.0 g of B. What
mass of product will be produced given that the
molar mass of A is 10.0 g/mol, B is
20.0 g/mol, and C is 25.0 g/mol? They react
according to the equation:
A + 3B  2C
Chapter 3 | Slide 41
Percent Yield
• An important indicator of the efficiency of
a particular laboratory or industrial
reaction.
Actual yield
 100%  percent yield
Theoretica l yield
3.10
Chapter 3 | Slide 42
For Review
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Stoichiometry
˙Deals with the amounts of substances consumed and/or produced in a chemical reaction.
˙We count atoms by measuring the mass of the sample.
˙To relate mass and the number of atoms, the average atomic mass is required.
Mole
˙The amount of carbon atoms in exactly 12 g of pure 12C.
˙6.022 x1023 units of a substance
˙The mass of one mole of an element= the atomic mass in grams
Molar mass
˙Mass(g) of one mole of a compound or element
˙Obtained for a compound by finding the sum of the average masses of its constituent atoms
Percent composition
˙The mass percent of each element in a compound
mass of element in 1 mole of substance
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Mass percent=
100%
mass of 1 mole of substance
Empirical formula
˙The simplest whole-number ratio of the various types of atoms in a compound
˙Can be obtained from the mass percent of elements in a compound
Molecular formula
˙For molecular substances:
˙The formula of the constituent molecules
˙Always an integer multiple of the empirical formula
˙For ionic substances:
Chapter 3 | Slide 43
˙The
same as the empirical formula
For Review
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Chemical reactions
˙Reactants are turned into products.
˙Atoms are neither created nor destroyed.
˙All of the atoms present in the reactants must also be present in the products.
Characteristics of a chemical equation
˙Represents a chemical reaction
˙Reactants on the left side of the arrow, products on the right side
˙When balanced, gives the relative numbers of reactant and product molecules or ions
Stoichiometry calculations
˙Amounts of reactants consumed and products formed can be determined from the
balanced chemical equation.
˙The limiting reactant is the one consumed first, thus limiting the amount of product that
can form.
Yield
˙The theoretical yield is the maximum amount that can be produced from a given
amount of the limiting reactant.
˙The actual yield, the amount of product actually obtained is always less than the
theoretical yield.
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actual yield(g)
P ercent
yield=
100%
Chapter
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theoretical yield(g)
Chapter Three
Stoichiometry
案例/討論
Figure 3.1 (left) A Scientist Injecting a
Sample into a Mass Spectrometer. (right)
Schematic Diagram of a Mass Spectrometer
Chapter 3 | Slide 46
Figure 3.2b Peaks
Chapter 3 | Slide 47
Figure 3.2c Bar Graph
Chapter 3 | Slide 48
Figure 3.3 Mass Spectrum of Natural
Copper
Chapter 3 | Slide 49
Juglone
胡桃醌
Chapter 3 | Slide 50
Isopentyl Acetate
乙酸異戊酯
Chapter 3 | Slide 51
Fullerene (Bucky ball)
Chapter 3 | Slide 52
Carvone
香芹酮
Chapter 3 | Slide 53
Carbon Dioxide
Chapter 3 | Slide 54
Water
Chapter 3 | Slide 55
Figure 3.5 A Schematic Diagram of the
Combustion Device Used to Analyze
Substances for Carbon and Hydrogen
Chapter 3 | Slide 56
Figure 3.6 Substances Whose
Empirical and Molecular Formulas Differ
Chapter 3 | Slide 57
Figure 3.9 Three Different Stoichiometric
Mixtures of Methane and Water, which
React One-to-One
Chapter 3 | Slide 58
Calculating Mass of Reactants and
Products
Chapter 3 | Slide 59
Figure 3.10 A Mixture of CH4 and H20
Molecules
Chapter 3 | Slide 60
Figure 3.11 Methane and Water Have
Reacted to Form Products
Chapter 3 | Slide 61
Figure 3.12 Hydrogen and Nitrogen
React to Form Ammonia
Chapter 3 | Slide 62
Solving a
Stoichiometry
Problem
Invovling
Masses of
Reactants
and Products
Chapter 3 | Slide 63
Jellybeans Can be Counted by
Weighing
Chapter 3 | Slide 64
Weighing Hex Nuts
Chapter 3 | Slide 65
Copper Nugget
Chapter 3 | Slide 66
Figure 3.4 Samples Containing One Mole
Each of Copper, Aluminum, Iron, Sulfur,
Iodine, and Mercury
Chapter 3 | Slide 67
Pure Aluminum
Chapter 3 | Slide 68
Bee Stings Cause the Release of
Isopentyl Acetate
乙酸異戊酯
Chapter 3 | Slide 69
Penicillin is Isolated from a Mold that Can be Grown in
Large Quantities in Fermentation Tanks
Chapter 3 | Slide 70
Figure 3.7 The Two Forms of
Dichloroenthane
Chapter 3 | Slide 71
Figure 3.8 The Structure of P4O10.
Chapter 3 | Slide 72
Computer-Generated Molecule of
Caffeine
Chapter 3 | Slide 73
Methane
Reacts with
Oxygen to
Produce
Flame
Chapter 3 | Slide 74
Hydrochloric
Acid Reacts
with Solid
Sodium
Hydrogen
Carbonate to
Produce
Gaseous
Carbon
Dioxide
Chapter 3 | Slide 75
Decomposition of Ammonium
Dichromate
Chapter 3 | Slide 76
Decomposition of Ammonium
Dichromate
Chapter 3 | Slide 77
Astronaut
Sidney M.
Gutierrez
Chapter 3 | Slide 78
Milk of
Magnesia
Chapter 3 | Slide 79
Race Cars use Methanol as a Fuel
Chapter 3 | Slide 80
Table 3.1 Comparison of 1 Mole
Samples of Various Elements
Chapter 3 | Slide 81
Table 3.2 Information Conveyed by the
Balanced Equation for the Combustion
of Methane
Chapter 3 | Slide 82
Chapter Three
Stoichiometry
問答
Question
• Indium has atomic number 49 and atomic
mass 114.8 g. Naturally occurring indium
contains a mixture of indium-112 and indium115, respectively, in an atomic ratio of
approximately
a)
b)
c)
d)
e)
6:94.
25:75.
50:50.
75:25.
94:6.
Chapter 3 | Slide 84
Answer
• a) 6:94.
• Section 3.2, Atomic Masses
• The ratio of x/(100 – x) may be obtained from
the following equation:
• (x/100)(112 amu) + [(100 – x)/100](115 amu)
•
= 114.8 amu
Chapter 3 | Slide 85
Question
• You have a sample of zinc (Zn) and a sample of
aluminum (Al). Each sample contains the same
number of atoms. Which of the following statements
concerning the masses of the samples is true?
– The mass of the zinc sample is more than twice as great as
the mass of the aluminum sample.
– The mass of the zinc sample is more than the mass of the
aluminum sample, but it is not twice as great.
– The mass of the aluminum sample is more than twice as
great as the mass of the zinc sample.
– The mass of the aluminum sample is more than the mass of
the zinc sample, but it is not twice as great.
– The masses of the two samples are equal.
Chapter 3 | Slide 86
Answer
•a)
The mass of the zinc sample is more than twice as
great as the mass of the aluminum sample.
•Section 3.2, Atomic Masses
•Zinc has an atomic mass more than twice that of
aluminum, so a sample of zinc with the same number of
atoms as a sample of aluminum will have more than twice
the mass of the aluminum.
Chapter 3 | Slide 87
Question
• Which of the following is the most accurate
description of a mole?
– The mass of carbon in a measured sample of carbon.
– The number of atoms in any given mass of an
element.
– The number of sodium ions in 58.44 g of sodium
chloride.
– At least two of the above are accurate descriptions of
a mole.
Chapter 3 | Slide 88
Answer
•c) The number of sodium ions in 58.44 g of
sodium chloride.
•Section 3.3, The Mole
•The molar mass of sodium chloride (NaCl) is
58.44 g. Because there is 1 atom of sodium for
each unit of sodium chloride, 58.44 g of NaCl
will contain 1 mol of sodium ions.
Chapter 3 | Slide 89
Question
• Which of the following is closest to the
average mass of one atom of copper?
– 63.55 g
– 52.00 g
– 58.93 g
– 65.38 g
– 1.055 × 10-22 g
Chapter 3 | Slide 90
Answer
•e) 1.055 × 10-22 g
•Section 3.3, The Mole
•1 atom of Cu × (63.55g/6.022 × 1023 atoms)
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= 1.055 × 10-22 g
•Also, one atom must have a very low mass,
and answer (e) is the only choice with the
correct order of magnitude.
Chapter 3 | Slide 91
Question
• Which of the following 100.0-g samples
contains the greatest number of atoms?
– Magnesium
– Zinc
– Silver
– Calcium
– All samples contains the same number of
atoms.
Chapter 3 | Slide 92
Answer
• a) Magnesium
• Section 3.3, The Mole
• Divide 100.0 g by each of the atomic masses.
• Because Mg has the lowest atomic mass, the
number of atoms must be the greatest.
Chapter 3 | Slide 93
Question
•For which of the following compounds does 1.0 g
represent 2.27  10–2 mol?
• a) H2O
• b) CO2
• c) NH3
• d) C2H6
•
Chapter 3 | Slide 94
Answer
• b) CO2
• Section 3.4, Molar Mass
• The answer may be obtained by solving for
molar mass:
• (1.0 g)/(molar mass) = 2.27  10–2 mol
• The molar mass obtained (44 g/mol) is closest to
the answer CO2.
Chapter 3 | Slide 95
Question
• The mass of 0.82 mol of a diatomic
molecule is 131.3 g. Identify the molecule.
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a) F2
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b) Cl2
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c) Br2
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d) I2
•
Chapter 3 | Slide 96
Answer
• c) Br2
• Section 3.4, Molar Mass
• The answer may be obtained by solving for
molar mass:
• (131.3 g)/(molar mass) = 0.82 mol
• The molar mass obtained (160 g/mol) is closest
to the answer Br2.
Chapter 3 | Slide 97
Question
• Which of the following 100.0-g samples
contains the greatest number of oxygen
atoms?
– H2O
– N2O
– C3H6O2
– CO2
– All of the samples have the same number
of oxygen atoms.
Chapter 3 | Slide 98
Answer
• a) H2O
• Section 3.4, Molar Mass
• Divide 100.0 g by the molar mass of each
compound and multiply by the number of oxygen
atoms in the formula.
Chapter 3 | Slide 99
Question
• Which of the following 100.0-g samples
contains the highest percent oxygen by
mass?
– H2O
– N2O
– C3H6O2
– CO2
– All of the samples have the same percent
oxygen by mass.
Chapter 3 | Slide 100
Answer (part 1)
• a) H2O
• Section 3.5, Percent Composition of Compounds
• The mass of the samples does not matter, nor
does it matter that the samples have the same
mass.
Chapter 3 | Slide 101
Answer (part 2)
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H2O: [(16.00 g)/(2.016 + 16.00)] × 100%
= 88.81% oxygen
N2O: [(16.00 g)/(28.02 + 16.00)] × 100%
= 36.35% oxygen
C3H6O2: [(32.00 g)/(36.03 + 6.048 + 32.00)]
× 100% = 43.20% oxygen
CO2: [(32.00 g)/(12.01 + 32.00)] × 100%
= 72.71% oxygen
Chapter 3 | Slide 102
Question
•The empirical formula of styrene is CH; its
molar mass is 104.1. What is the molecular
formula of styrene?
• a) C2H4
• b) C8H8
• c) C10H10
• d) C6H6
•
Chapter 3 | Slide 103
Answer
• b) C8H8
• Section 3.6, Determining the Formula of a
Compound
• The molecular formula is (CH)n, so n may be
obtained from the following equation:
• (n)(12.0 g/mol + 1.0 g/mol) = 104.1 g/mol
Chapter 3 | Slide 104
Question
•When the equation
•NH3 + O2  NO + H2O
•is balanced with the smallest set of integers, the
sum of the coefficients is
• a) 4
• b) 12
• c) 14
• d) 19
• e) 24
Chapter 3 | Slide 105
Answer
• d) 19
• Section 3.8, Balancing Chemical Equations
• To have the same number of each atom on
each side of the equation, we get
• 4NH3 + 5O2  4NO + 6H2O
• The sum of the coefficients is 19.
Chapter 3 | Slide 106
Question
• How many of the following correctly
balance this chemical equation:
• CaO + C  CaC2 + CO2
i. CaO2 + 3C  CaC2 + CO2
ii. 2CaO + 5C  2CaC2 + CO2
iii. CaO + (5/2)C  CaC2 + (1/2)CO2
iv. 4CaO + 10C  4CaC2 + 2CO2
a) 0
Chapter 3 | Slide 107
b) 1
c) 2
d) 3
e) 4
Answer
•d) 3
•Section 3.8, Balancing Chemical Equations
•Choices ii, iii, and iv are all correct, although only
choice ii is written in standard form.
Chapter 3 | Slide 108
Question
• How many of the following statements
are true concerning balanced chemical
equations?
i. The number of molecules is conserved.
ii. Coefficients indicate mass ratios of the
substances involved.
iii. Atoms are neither created nor destroyed.
iv. The sum of the coefficients on the left side
equals the sum of the coefficients on the
right side.
a) 0
Chapter 3 | Slide 109
b) 1
c) 2
d) 3
e) 4
Answer
• b) 1
• Section 3.8, Balancing Chemical
Equations
• Choice iii is the only correct choice.
Chapter 3 | Slide 110
Question (part 1)
• In the reaction
• 2A + 3B  C
• 4.0 moles of A reacts with 4.0 moles of B.
• Which of the following choices best answers the
question: “Which reactant is limiting?”
Chapter 3 | Slide 111
Question (part 2)
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a)
Neither is limiting because equal
amounts (4.0 mol) of each reactant are
used.
b)
A is limiting because 2 is smaller
than 3 (the coefficients in the balanced
equation).
c)
A is limiting because 2 mol is
available but 4.0 mol is needed.
d)
B is limiting because 3 is larger than
2 (the coefficients in the balanced
equation).
e)
B is limiting because 4.0 mol is
available but 6.0 mol is needed.
Chapter 3 | Slide 112
Answer
•e)
B is limiting because 4.0 mol is available but 6.0 mol
is needed.
•Section 3.10, Calculations Involving a Limiting Reactant
•According to the balanced chemical equation, 4.0 mol of A
would require
•(4.0 mol A)(3 mol B/2 mol A) = 6.0 mol B
•but only 4.0 mol of B is available. B is the limiting reactant.
Chapter 3 | Slide 113
Question
•
The limiting reactant in a reaction
•
a) has the smallest coefficient in a balanced
equation.
• b) is the reactant for which you have the
fewest number of moles.
c) has the lowest ratio of [moles available/
•
coefficient in the balanced equation].
• d) has the lowest ratio of [coefficient in the
balanced equation/moles available].
• e) none of these
Chapter 3 | Slide 114
Answer (part 1)
•c) has the lowest ratio of [moles available/
• coefficient in the balanced equation].
•Section 3.10, Calculations Involving a Limiting
Reactant
Chapter 3 | Slide 115
Answer (part 2)
• Using the moles of materials available and the
balanced chemical equation, the ratio of the
moles of a material available to its
corresponding coefficient in the balanced
equation gives the quantity available per
quantity needed. In comparing these ratios, the
smallest will be the limiting reactant.
Chapter 3 | Slide 116
Question (part 1)
• Consider the following balanced equation:
• A + 5B  3C + 4D
• Which of the following choices best
answers the question: “When equal
masses of A and B are reacted, which is
limiting?”
Chapter 3 | Slide 117
Question (part 2)
• a) If the molar mass of A is greater than the
molar mass of B, then A must be limiting.
• b) If the molar mass of A is less than the molar
mass of B, then A must be limiting.
• c) If the molar mass of A is greater than the
molar mass of B, then B must be limiting.
• d) If the molar mass of A is less than the molar
mass of B, then B must be limiting.
Chapter 3 | Slide 118
Answer
•d) If the molar mass of A is less than the molar mass of
B, then B must be limiting.
•Section 3.10, Calculations Involving a Limiting Reactant
•Because the masses are equal, if the molar mass of A is
less than the molar mass of B, the number of moles of A
will be greater than the number of moles of B. Given that
more moles of B are required (a 1:5 ratio of A:B according
to the balanced equation), B must be limiting.
Chapter 3 | Slide 119
Question
• Which of the following reaction mixtures
would produce the greatest amount of
product according to the following chemical
equation:
• 2H2 + O2  2H2O
–
–
–
–
–
2 mol H2 and 2 mol O2
2 mol H2 and 3 mol O2
2 mol H2 and 1 mol O2
3 mol H2 and 1 mol O2
All of these choices would produce the same
amount of product.
Chapter 3 | Slide 120
Answer (part 1)
•e) All of these choices would produce the same
amount of product.
•Section 3.10, Calculations Involving a Limiting
Reactant
•Each combination will produce 2 moles of H2O.
Chapter 3 | Slide 121
Answer (part 2)
• a) H2 is limiting: 2.0 mol H2 will produce 2.0 mol H2O.
• b) H2 is limiting: 2.0 mol H2 will produce 2.0 mol H2O.
• c) Neither is limiting: 2.0 mol H2 will produce 2.0 mol H2O;
1.0 mol O2 will produce 2.0 mol H2O.
• d) O2 is limiting: 1.0 mol O2 will produce 2.0 mol H2O.
Chapter 3 | Slide 122
第三章習題中的一些分子
51. Ascorbic acid (Vitamin C)
抗壞血酸(維他命C)
60.Anabolic steroid (合成代謝類固醇,
甾類同化激素 )
63. Vitamin B12 （cyanocobalamin）
維他命B12（氰鈷素）
64.Fugal laccase (真菌漆酶 )
95 Aspirin（阿司匹林）
102. Acrylonitrile (丙烯腈)
112 Terephthalic acid (對苯二甲酸)
142. Tetrodotoxin (河魨素)