Chemical Equations: Quantities of Reactants

1
Chapter 4 Quantities of Reactants Products
General Chemistry I Talia Ara
2
A. Chemical Equations (Reactions)

• Understanding chemical reactions is one of the
  fundamental pursuits of chemistry.
• Chemical reaction a process
• by which substances change
• into other substances by
• rearrangement, combination,
• or separation of atoms
• eg. A burning candle is an example
• of a combustion reaction.

3
A. Chemical Equations (Reactions)

• A chemical equation is the symbolic
  representation of a chemical reaction in terms of
  chemical formulas.
• 4 Fe (s) 3 O2 (g) 2 Fe2O3 (s)

eg. The reaction of powdered Iron with oxygen to
form iron oxide is represented by the equation
above.
4
1. Writing a Chemical Equation

• a) Reactants Products In a chemical equation,
  the reactants (starting materials) are listed on
  the left the products are listed on the right.
• The reaction proceeds from left to right.
• Reactants Products
• (In words, the arrow means yields)

5
1. Writing a Chemical Equation

• b) Physical States the physical states of the
  reactants and products in the reaction are
  indicated in parentheses after each chemical
  formula
• 2 Na (s) Cl2 (g) 2 NaCl (s)
• (s) solid, (l) liquid, (g) gas,
• (aq) aqueous solution

6
1. Writing a Chemical Equation

• c) Coefficients the relative amounts of each
  substance in the reaction are indicated with
  stoichiometric coefficients
• 2 Na (s) Cl2 (g) 2 NaCl (s)
• eg. Notice there are the same number of Na atoms
  on each side of the equation, and the same number
  of Cl atoms.

7
2. Types of Chemical Reactions

• There are far too many chemical reactions to
  memorize each and every one.
• Many reactions fall into one of the simple
  reaction patterns shown below

8
2. Types of Chemical Reactions

• a) Combination Reaction reaction in which two or
  more substances combine to form one product
• Oxygen and the halogens (Group 7A) frequently
  undergo combination reactions.

9
a) Combination Reaction
Combination reaction between zinc and iodine

Zn (s) I2 (s) ZnI2 (s)
10
2. Types of Chemical Reactions

• b) Decomposition Reaction reaction in which one
  substance decomposes to form two or more products
  (the opposite of a combination reaction)
• Some compounds that are stable under normal
  conditions will decompose upon heating.

11
b) Decomposition Reaction
Decomposition of nitroglycerine

• 4 C3H5(NO3) 3 (l) 12 CO2 (g) 10 H2O (g)
  6 N2 (g) O2 (g)
• Nitroglycerine is very sensitive to heat, light
  and shock.
• It can decompose violently.

12
2. Types of Chemical Reactions

• c) Displacement Reaction reaction in which one
  element reacts with a compound to form a new
  compound and release a different element
• The element released is said to have been
  displaced.

13
c) Displacement Reaction
Displacing H with Na The Reaction of Sodium
Metal with Water

14
2. Types of Chemical Reactions

• d) Exchange Reaction reaction in which there is
  an interchange of partners between two compounds
• Exchange reactions commonly occur between ionic
  compounds dissolved in water (more in the next
  chapter).

15
d) Exchange Reaction
Exchange Reaction Between Lead Nitrate and
Potassium Chromate

Pb(NO3)2 (aq) K2CrO4 (aq) PbCrO4
(s) 2 KNO3 (aq) lead nitrate
potassium chromate lead chromate
potassium nitrate
16
B. Balancing Chemical Equations

• The Law of Conservation of Mass dictates that the
  total number of atoms of each element on each
  side of a chemical equation must match.
• - An equation that meets this criterion is said
  to be a balanced equation.

17
B. Balancing Chemical Equations

• For example
• CaCO3 (s) CaO (s) CO2 (g)
• Reactants Products
• 1 Ca atom 1 Ca atom
• 1 C atom 1 C atom
• 3 O atoms 3 O atoms

18
2. Rules for Balancing Chemical Equations
Step 1 Write an unbalanced equation containing
the correct formulas of all the reactants and
products.
19
2. Rules for Balancing Chemical Equations

• Step 1 Write an unbalanced equation containing
  the correct formulas of all the reactants and
  products.
• eg. Ammonia from hydrogen and nitrogen
• H2 (g) N2 (g) NH3 (g)

20
2. Rules for Balancing Chemical Equations

• Step 2 Use a whole-integer coefficient to
  balance the atoms of one of the elements.
• Which element should you start with?
• Start with the element that appears in the fewest
  formulas preferably one that does not appear
  more than once on the same side of the equation.
• Do not start with an element that is already
  balanced.

21
2. Rules for Balancing Chemical Equations
Step 2 Use a whole-integer coefficient to
balance the atoms of one of the elements.
H2 (g) N2 (g) NH3 (g)
22
2. Rules for Balancing Chemical Equations
Step 2 Use a whole-integer coefficient to
balance the atoms of one of the elements.
H2 (g) N2 (g) 2 NH3 (g)
-Adding a 2 to NH3 balances the nitrogen atoms
at two per side. - A coefficient multiplies all
atoms in the corresponding formula there are 2
N atoms and 6 H atoms on the right hand side of
the equation.
23
2. Rules for Balancing Chemical Equations
Step 3 Balance the atoms of all the remaining
elements. This may involve some trial and error
in more complicated reactions. H2 (g) N2
(g) 2 NH3 (g)
24
2. Rules for Balancing Chemical Equations

• Step 3 Balance the atoms of all the remaining
  elements. This may involve some trial and error
  in more complicated reactions.
• 3 H2 (g) N2 (g) 2 NH3 (g)
• Adding a 3 to H2 brings the total number of
• hydrogen atoms per side to six.

25
2. Rules for Balancing Chemical Equations
Step 4 Verify that the number of atoms of each
element is balanced. 3 H2 (g) N2 (g) 2
NH3 (g)
26
2. Rules for Balancing Chemical Equations
Step 4 Verify that the number of atoms of each
element is balanced. 3 H2 (g) N2 (g) 2
NH3 (g) (3 x 2)H 2 N 2 N (2 x 3)H 6 H
2 N 2 N 6 H Balanced!
27
Some Practice
S8 (s) F2 (g) SF6 (g)
28
Some Practice
S8 (s) F2 (g) SF6 (g) S8 (s)
F2 (g) 8 SF6 (g)
29
Some Practice
S8 (s) F2 (g) SF6 (g) S8 (s)
F2 (g) 8 SF6 (g) S8 (s) 24 F2
(g) 8 SF6 (g)
30
Some Practice

• S8 (s) F2 (g) SF6 (g)
• S8 (s) F2 (g) 8 SF6 (g)
• S8 (s) 24 F2 (g) 8 SF6 (g)
• 8 S (24 x 2)F 8 S (8 x 6)F
• 8 S 48 F 8 S 48 F
• Balanced!

31
Some Practice
Al(OH)3 (s) Al2O3 (s) H2O (g)
32
Some Practice
Al(OH)3 (s) Al2O3 (s) H2O (g) 2 Al(OH)3
(s) Al2O3 (s) H2O (g)
33
Some Practice
Al(OH)3 (s) Al2O3 (s) H2O (g) 2 Al(OH)3
(s) Al2O3 (s) H2O (g) 2 Al(OH)3 (s)
Al2O3 (s) 3 H2O (g)
34
Some Practice
Al(OH)3 (s) Al2O3 (s) H2O (g) 2 Al(OH)3
(s) Al2O3 (s) H2O (g) 2 Al(OH)3 (s)
Al2O3 (s) 3 H2O (g) 2 Al (2 x 3)O (2
x 3)H 2 Al (3 3)O (3 x 2) H 2 Al
6 O 6 H 2 Al 6 O 6 H Balanced!
35
Some Practice
Fe (s) Cl2 (g) FeCl3 (s)
36
Some Practice
Fe (s) Cl2 (g) FeCl3 (s) Fe (s)
3 Cl2 (g) 2 FeCl3 (s)
37
Some Practice
Fe (s) Cl2 (g) FeCl3 (s) Fe (s)
3 Cl2 (g) 2 FeCl3 (s) 2 Fe (s) 3 Cl2
(g) 2 FeCl3 (s)
38
Some Practice
Fe (s) Cl2 (g) FeCl3 (s) Fe (s)
3 Cl2 (g) 2 FeCl3 (s) 2 Fe (s) 3 Cl2
(g) 2 FeCl3 (s) 2 Fe (3 x 2)Cl 2 Fe
(2 x 3) Cl 2 Fe 6 Cl 2 Fe 6 Cl
Balanced!
39
Some Practice
C5H12 (g) O2 (g) CO2 (g) H2O (g)
40
Some Practice
C5H12 (g) O2 (g) CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) H2O
(g)
41
Some Practice
C5H12 (g) O2 (g) CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) 6 H2O
(g)
42
Some Practice
C5H12 (g) O2 (g) CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) 6 H2O
(g) C5H12 (g) 8 O2 (g) 5 CO2 (g) 6 H2O
(g)
43
Some Practice
C5H12 (g) O2 (g) CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) H2O
(g) C5H12 (g) O2 (g) 5 CO2 (g) 6 H2O
(g) C5H12 (g) 8 O2 (g) 5 CO2 (g) 6 H2O
(g) 5 C 12 H (8 x 2) O 5 C (6 x
2) H (10 6) O 5 C 12 H 16 O 5 C
12 H 16 O Balanced!
44
Some Practice
Ni(OH)2 (aq) HNO3 (aq) Ni(NO3)2 (aq) H2O
(l) Try this one at home
45
C. Stoichiometry

• Stoichiometry the study of the quantitative
  relationships between amounts of reactants and
  products in a chemical reaction
• Stoichiometric Coefficients the multiplying
  numbers assigned to the species in a chemical
  equation in order to balance the equation
• CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)

46
1. The Macro-Nano Connection

• On the nanoscale, the stoichiometric coefficients
  in an equation represent the relative numbers of
  molecules or atoms in a reaction.
• CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)

1 molecule
2 molecules
2 molecules
1 molecule
47
1. The Macro-Nano Connection

• On the macroscale, the coefficients represent the
  molar ratios, the relative number of moles of
  reactants and products in a reaction.
• These molar ratios can be used to calculate the
  molar amount of one compound from the molar
  amount of another.
• CH4 (g) 2 O2 (g) CO2 (g) 2 H2O (g)
• 2 mol O2 1 mol CO2 2 mol H2O
• 1 mol CH4 1 mol CH4 1 mol CO2

48
1. The Macro-Nano Connection

• By incorporating the molar masses of the
  compounds, the relative masses of the reactants
  and products in a balanced chemical equation can
  be calculated.
• Notice the total mass of the reactants must equal
  the total mass of the products in a balanced
  equation.

49
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) Using this
balanced equation, determine the number of moles
of NH3 that could be obtained from 3.6 mol H2.
50
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) The
coefficients in the balanced equation represent
the molar ratios, so
51
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) The
coefficients in the balanced equation represent
the molar ratios, so 2 mol NH3 form per 3 mol
H2 reacted
52
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) The
coefficients in the balanced equation represent
the molar ratios, so 2 mol NH3 form per 3 mol
H2 reacted 3.6 mol H2
53
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) The
coefficients in the balanced equation represent
the molar ratios, so 2 mol NH3 form per 3 mol
H2 reacted 3.6 mol H2 x 2 mol NH3
3 mol H2
54
a) Using Molar Ratios
N2 (g) 3 H2 (g) 2 NH3 (g) The
coefficients in the balanced equation represent
the molar ratios, so 2 mol NH3 form per 3 mol
H2 reacted 3.6 mol H2 x 2 mol NH3
2.4 mol NH3 3 mol H2
55
a) Using Molar Ratios
Determine the number of moles of N2 required to
react completely with 4.5 mol H2. N2 (g) 3
H2 (g) 2 NH3 (g)
56
a) Using Molar Ratios
Determine the number of moles of N2 required to
react completely with 4.5 mol H2. N2 (g) 3
H2 (g) 2 NH3 (g) 1 mol N2 per 3 mol H2
57
a) Using Molar Ratios
Determine the number of moles of N2 required to
react completely with 4.5 mol H2. N2 (g) 3
H2 (g) 2 NH3 (g) 1 mol N2 per 3 mol H2
4.5 mol H2
58
a) Using Molar Ratios
Determine the number of moles of N2 required to
react completely with 4.5 mol H2. N2 (g) 3
H2 (g) 2 NH3 (g) 1 mol N2 per 3 mol H2
4.5 mol H2 x 1 mol N2 3 mol H2
59
a) Using Molar Ratios
Determine the number of moles of N2 required to
react completely with 4.5 mol H2. N2 (g) 3
H2 (g) 2 NH3 (g) 1 mol N2 per 3 mol H2
4.5 mol H2 x 1 mol N2 1.5 mol N2
3 mol H2
60
2. Mass Relationships

• To relate masses of reactants and products, molar
  masses must be used in addition to molar ratios.
• Dividing a mass amount by the molar mass converts
  it to moles.
• Multiplying a molar amount by the molar mass
  converts it to grams.

61
2. Mass Relationships
For any two chemical substances in a balanced
equation
Grams A Moles A (divide by molar mass
A) (multiply by molar
ratio) Grams B Moles B (multiply
by molar mass B)
62
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 2.5 mol N2. N2 (g) 3 H2
(g) 2 NH3 (g)
63
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 2.5 mol N2. N2 (g) 3 H2
(g) 2 NH3 (g) Moles of N2 Moles of NH3 Grams
of NH3
64
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 2.5 mol N2. N2 (g) 3 H2
(g) 2 NH3 (g) Moles of N2 Moles of NH3 Grams
of NH3 Molar Ratio 2 mol NH3 formed per 1 mol
N2 reacted Molar Mass of NH3 17.03 g/mol
65
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 2.5 mol N2. N2 (g) 3 H2
(g) 2 NH3 (g) Moles of N2 Moles of NH3 Grams
of NH3 Molar Ratio 2 mol NH3 formed per 1 mol
N2 reacted Molar Mass of NH3 17.03 g/mol 2.5
mol N2 x 2 mol NH3 1 mol N2
66
2. Mass Relationships

• Determine the mass (in grams) of NH3 that could
  be obtained from 2.5 mol N2.
• N2 (g) 3 H2 (g) 2 NH3 (g)
• Moles of N2 Moles of NH3 Grams of NH3
• Molar Ratio 2 mol NH3 formed per 1 mol N2
  reacted
• Molar Mass of NH3 17.03 g/mol
• 2.5 mol N2 x 2 mol NH3 x 17.03 g
  85 g NH3
• 1 mol N2 1 mol NH3

67
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g)
68
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g) Grams H2 Moles H2 Moles NH3
Grams NH3
69
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g) Grams H2 Moles H2 Moles NH3
Grams NH3 Molar Mass H2 2.02 g/mol Molar
Ratio 2 mol NH3 formed per 3 mol H2
reacted Molar Mass NH3 17.03 g/mol
70
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g) Grams H2 Moles H2 Moles NH3
Grams NH3 Molar Mass H2 2.02 g/mol Molar
Ratio 2 mol NH3 formed per 3 mol H2
reacted Molar Mass NH3 17.03 g/mol 3.00 g H2
x 1 mol H2 2.02 g H2
71
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g) Grams H2 Moles H2 Moles NH3
Grams NH3 Molar Mass H2 2.02 g/mol Molar
Ratio 2 mol NH3 formed per 3 mol H2
reacted Molar Mass NH3 17.03 g/mol 3.00 g H2
x 1 mol H2 x 2 mol NH3
2.02 g H2 3 mol H2
72
2. Mass Relationships
Determine the mass (in grams) of NH3 that could
be obtained from 3.00 g H2. N2 (g) 3 H2
(g) 2 NH3 (g) Grams H2 Moles H2 Moles NH3
Grams NH3 Molar Mass H2 2.02 g/mol Molar
Ratio 2 mol NH3 formed per 3 mol H2
reacted Molar Mass NH3 17.03 g/mol 3.00 g H2
x 1 mol H2 x 2 mol NH3 x 17.03 g
2.02 g H2 3 mol H2
1 mol NH3 16.9 g NH3
73
3. Limiting Reactant (Reagent)
When the reactants are present in
stoichiometric amounts, the same molar ratios as
indicated in the balanced equation, they can be
consumed entirely as the reaction goes to
completion. eg. CO 2 H2 CH3OH
1 mol CO 2 mol H2 1 mol
CH3OH 28 g CO 4 g H2
32 g CH3OH
74
3. Limiting Reactant (Reagent)

• But in many reactions the relative amounts of
  the reactants are unbalanced (not
  stoichiometric).
• How do you analyze a reaction like that?
• For example, if you mix 1 mole of CO and 1 mole
  of H2, how much methanol should you expect to
  obtain?
• CO 2 H2 CH3OH
• 1 mol CO 1 mol H2 ???

75
3. Limiting Reactant (Reagent)

• In those cases, one reactant is present in
  excess, and the other is said to be the limiting
  reactant.
• Limiting reactant the reactant that is
  completely converted to products in a reaction
• Once the limiting reactant has been used up, the
  reaction stops, and no more product can form.

76
3. Limiting Reactant (Reagent)

• The limiting reactant ultimately determines how
  much product can be obtained in a chemical
  reaction. (eg. making cheese sandwiches)

77
3. Limiting Reactant
If 2.0 moles of nitrogen and 3.0 moles of
hydrogen are used in the following reaction, how
many moles of NH3 would be formed? N2 (g)
3 H2 (g) 2 NH3 (g)
78
3. Limiting Reactant
If 2.00 moles of nitrogen and 3.00 moles of
hydrogen are used in the following reaction, how
many grams of NH3 would be formed? N2 (g)
3 H2 (g) 2 NH3 (g) One way to figure this
out - Start by calculating the mass of product
(NH3) expected from each given quantity of
reactant. - The smallest value will be
associated with the limiting reactant, and will
tell you how much product to expect.
79
3. Limiting Reactant
How many grams of NH3 are formed in the following
reaction? N2 (g) 3 H2 (g) 2 NH3 (g)
(2.00 mol N2) (3.00 mol H2) ???
80
3. Limiting Reactant
How many grams of NH3 are formed in the following
reaction? N2 (g) 3 H2 (g) 2 NH3 (g)
(2.00 mol N2) (3.00 mol H2)
??? 2.00 mol N2 x 2 mol NH3 x 17.03 g
68.1 g NH3 1 mol N2
1 mol NH3 3.00 mol H2 x 2 mol NH3
x 17.03 g 34.1 g NH3 3
mol H2 1 mol NH3 -Hydrogen is the
limiting reagent, and 34.1 g of NH3 can be form
in this reaction.
81
3. Limiting Reactant

• N2 (g) 3 H2 (g) 2 NH3 (g)
• 2.00 mol N2 3.00 mol H2 34.1 g NH3
• If hydrogen is the limiting reactant, how many
  grams of nitrogen (N2) will remain unreacted at
  the end of the reaction?
• Start by calculating how much N2 is used up in
  the reaction.
• You can then calculate how much is left over
  after the limiting reactant is used up.

82
3. Limiting Reactant
N2 (g) 3 H2 (g) 2 NH3 (g) 2.00
mol N2 3.00 mol H2 34.1 g NH3

83
3. Limiting Reactant
N2 (g) 3 H2 (g) 2 NH3 (g) 2.00
mol N2 3.00 mol H2 34.1 g NH3 3.00
mol H2 x 1 mol N2 1.00 mol N2
reacted 3 mol H2 2.00 mol N2 (start)
1.00 mol N2 (reacted) 1.00 mol N2
unreacted 1.00 mol N2 x 28.01 g
28.0 g N2 unreacted 1 mol N2
84
4. Theoretical Yield

• The maximum possible quantity of product that can
  be formed in a reaction, based on the amount of
  the limiting reactant at the beginning, is called
  the theoretical yield.
• Given a perfect set of reaction conditions and a
  reaction that proceeds all the way to completion,
  this is the maximum amount of product that can be
  expected.
• Not all reactions provide 100 of the theoretical
  yield.

85
4. Theoretical Yield

• The actual yield is the amount of product
  obtained in the laboratory when the reaction is
  run.
• The actual yield is often less than the
  theoretical yield.
• eg. If you heat 20 kernels of popping corn in a
  pan, you expect to get 20 pieces of popped corn.
• Theoretical Yield 20

86
4. Theoretical Yield

• In this case, the actual yield is 16 popped
  kernels (4 less than the theoretical yield of
  20).

87
5. Percent Yield
yield actual yield x 100
theoretical yield
88
5. Percent Yield

• yield actual yield x 100
• theoretical yield

yield (16/20) x 100 80
89
5. Percent Yield

• The percent yield of a reaction is a measure of
  the efficiency, or success, of that reaction.
• If the yield is close to 100, the reaction is
  very efficient (most of the reactants are
  converted to products).
• If the yield is closer to 0, the reaction is
  very inefficient (very little of the desired
  product is formed).
• The actual and percent yields are experimentally
  determined values.

90

• Hydrazine, a type of rocket fuel, is produced by
  the reaction of Cl2 with excess NaOH and NH3.
• 2 NaOH Cl2 2 NH3 N2H4 2 NaCl 2 H2O
• Calculate the theoretical yield of N2H4 (molar
  mass 32.045 g/mol) if 100.0 g of Cl2 (molar
  mass 70.906 g/mol) is reacted with excess NaOH
  and NH3.
• b) If 35.6 g of hydrazine are produced in this
  reaction, what is the percent yield?

91
a) Calculate the theoretical yield of N2H4 (molar
mass 32.045 g/mol) if 100.0 g of Cl2 (molar
mass 70.906 g/mol) is reacted with excess NaOH
and NH3.
92
a) Calculate the theoretical yield of N2H4 (molar
mass 32.045 g/mol) if 100.0 g of Cl2 (molar
mass 70.906 g/mol) is reacted with excess NaOH
and NH3. 100.0 g Cl2 x 1 mol 1.410
mol Cl2 70.906 g 1.410 mol Cl2 x 1
mol N2H4 1.410 mol N2H4
1 mol Cl2 1.410 mol N2H4 x
32.045 g 45.18 g N2H4 1 mol
93
b) If 35.6 g of hydrazine are produced in this
reaction, what is the percent yield?
94
b) If 35.6 g of hydrazine are produced in this
reaction, what is the percent yield? Actual
Yield 35.6 g Theoretical Yield 45.18
g Percent Yield (35.6 g / 45.18 g) x 100
78.8
95
CS2 (g) 3 O2 (g) CO2 (g) 2 SO2 (g)
76.143 g/mol 31.998 g/mol 44.009
g/mol 64.064 g/mol If the reaction starts
with 30.0 g of CS2 and 35.0 g of O2, what is the
theoretical yield of SO2? Which reactant is the
limiting reactant? If 38.5 g of SO2 form in the
reaction, what is the percent yield?
96
CS2 (g) 3 O2 (g) CO2 (g) 2 SO2 (g)
76.143 g/mol 31.998 g/mol 44.009
g/mol 64.064 g/mol
97
CS2 (g) 3 O2 (g) CO2 (g) 2 SO2
(g) CS2 76.143 g/mol O2 31.998 g/mol CO2
44.009 g/mol SO2 64.064 g/mol CS2
30.0g/76.143 g/mol 0.3940 mol CS2 (0.3940
mol CS2)(2 mol SO2/1 mol CS2) 0.7880 mol
SO2 (0.7880 mol SO2)(64.064 g/mol) 50.5 g
SO2 (CS2 excess) O2 35.0g/31.998 g/mol 1.094
mol O2 (1.094 mol O2)(2 mol SO2/3 mol O2)
0.7292 mol SO2 (0.7292 mol SO2)(64.064 g/mol)
46.7 g SO2 (O2 limiting) Percent Yield (38.5
g/46.7 g) x 100 82.4 82