Transcript CHM 111 CHAPTER 3-B Stoichiometry © 2012 by W. W. Norton & Company.

CHM 111
CHAPTER 3-B
Stoichiometry
© 2012 by W. W. Norton & Company
Avogadro and the Mole
2
Stoichiometry
•
Stoichiometry: Relates the moles of products and
reactants to each other and to measurable
quantities.
3
Stoichiometry
Aqueous solutions of NaOCl (household bleach)
are prepared by the reaction of NaOH with Cl2:
2 NaOH(aq) + Cl2(g)  NaOCl(aq) + NaCl(aq) + H2O(l)
How many grams of NaOH are needed to react
with 25.0 g of Cl2?
4
Stoichiometry
2 NaOH + Cl2 → NaOCl + NaCl + H2O
25.0 g Cl2 reacts with ? g NaOH
25.0 g Cl2 
1 m oleCl2
 0.353m olesCl2
70.90 g Cl2
1m oleCl2 2 m olesNaOH 40.0 g NaOH
25.0 g Cl2 


 28.2 g NaOH
70.90 g Cl2
1m oleCl2
1m oleNaOH
5
Avogadro and the Mole
•
Calculate the molar mass of the following:
Fe2O3 (Rust)
C6H8O7 (Citric acid)
C16H18N2O4 (Penicillin G)
•
Balance the following, and determine how many
moles of CO will react with 0.500 moles of Fe2O3.
Fe2O3(s) + CO(g)
Fe(s) + CO2(g)
6
Avogadro and the Mole
Fe2O3 + CO → Fe + CO2
Balance (not a simple one)
Save Fe for last
C is balanced, but can’t balance O
In the products the ratio C:O is 1:2 and can’t change
Make the ratio C:O in reactants 1:2
Fe2O3 + 3CO → 2Fe + 3CO2
7
Avogadro and the Mole
Fe2O3 + 3CO → 2Fe + 3CO2
3 m oleCO
0.500 m oles Fe2O3 
 1.50m olesCO
1m ole Fe2O3
8
Stoichiometry
9
Stoichiometry
•
Aspirin is prepared by reaction of salicylic acid
(C7H6O3) with acetic anhydride (C4H6O3) to form aspirin
(C9H8O4) and acetic acid (CH3CO2H). Use this
information to determine the mass of acetic anhydride
required to react with 4.50 g of salicylic acid. How many
grams of aspirin will result? How many grams of acetic
acid will be produced as a by-product?
10
Stoichiometry
Salicylic acid + Acetic anhydride →
Aspirin + acetic acid
C7H6O3 + C4H6O3 → C9H8O4 + CH3CO2H
C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2
Balanced!
Equal # moles for all
11
Stoichiometry
4.50 g Salicylic acid (C7H6O3) = ? moles
MW C7H6O3 = 7 x 12.01 + 6 x 1.008 + 3 x 16.00
= 138.12 g/mole
1m ole S. A.
4.50 g S. A. 
 0.0326m oles S. A.
138.12 g S. A.
12
Stoichiometry
Since all compounds have the same S.C., there must be
0.0326 moles of all 4 of them involved in the reaction.
g Aspirin (C9H8O4) = 0.0326 moles x MW Aspirin
= .0326 x [9x12.01 + 8x1.008 + 4x16.00]
=.0326 mole x 180.15 g/mole
5.87 g Aspirin
13
Stoichiometry
•
Yields of Chemical Reactions: If the actual amount
of product formed in a reaction is less than the
theoretical amount, we can calculate a percentage
yield.
Actual product yield
% yield 
 100%
Theoretical product yield
14
Stoichiometry
•
Dichloromethane (CH2Cl2) is prepared by reaction
of methane (CH4) with chlorine (Cl2) giving
hydrogen chloride as a by-product. How many
grams of dichloromethane result from the reaction
of 1.85 kg of methane if the yield is 43.1%?
15
Stoichiometry
CH4 + Cl2 → CH2Cl2 + HCl
Balance
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
16
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
1.85 kg CH4 = ? moles CH4
MW CH4 = 1x12.01 + 4x1.008 = 16.04 g/mole
1000g 1 m oleCH 4
1.85 kg CH 4 

 115m olesCH 4
kg
16.4 g CH 4
17
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
115 moles CH4
in theory we should produce:
115 moles of CH2Cl2 and 230 moles of HCl
And use up 230 moles of Cl2
18
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
115 moles of CH2Cl2 = ? g
MW CH2Cl2 = 12.01 + 2x1.008 + 2x35.45 = 84.93
115 moles x (84.03 g/mole) = 9770 g
19
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Expect 9770 g CH2Cl2
but the yield is 43.1%
So we produced just 0.431 x 9770 g
4.21 kg CH2Cl2
20
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Suppose the reaction went to completion
(100% yield)
Is mass conserved?
21
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
Start with 115 moles CH4 and 230 moles Cl2
total mass = 115x16.04 + 230x70.90
= 1850 + 16300 = 18150
only 3 sig. figs. → 18.2 kg
22
Stoichiometry
CH4 + 2Cl2 → CH2Cl2 + 2HCl
End with 115 moles CH2Cl2 and 230 moles HCl
total mass = 115x84.93 + 230x36.46
= 9770 + 8390 = 18160
only 3 sig. figs → 18.2 kg
23
Stoichiometry
•
Limiting Reagents:
The extent to which
a reaction takes
place depends on
the reactant that is
present in limiting
amounts—the
limiting reagent.
24
Limiting Reagent Video
25
Stoichiometry
•
In this figure, ethylene oxide is the limiting reagent.
26
Stoichiometry
Limiting Reagent: The reactant that governs the maximum
amount of product that can be formed.
Ex: Combustion of Octane
C8H18 + O2 → CO2 + H2O
If 100 g octane and 150 g oxygen are supplied for
combustion, which is the limiting reagent?
27
Stoichiometry
Limiting Reagent: The reactant that governs the maximum
amount of product that can be formed.
Ex: Combustion of Octane
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
If 100 g octane and 150 g oxygen are supplied for
combustion, which is the limiting reagent?
28
Stoichiometry
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
Moles octane: 100 g x mole/114.22 g = 0.876 moles
Moles oxygen: 150 g x mole/32.00 g = 4.69 moles
Moles oxygen needed: 0.876 x 25/2 = 11.0 moles
Moles octane needed: 4.69 x 2/25 = 0.375 moles
29
Stoichiometry
2 C8H18 + 25 O2 → 16 CO2 + 18 H2O
0.976 moles octane available and 0.375 moles needed
4.69 moles oxygen available and 11.0 moles needed
So OXYGEN is the limiting reagent
We won’t burn all of the octane
30
Some Interesting Chemical Reactions
1. Production of Smog
N2 + O2 + heat → 2 NO
2 NO + O2 → 2 NO2 (brown gas)
2. The Greenhouse Effect
2 C8H18 + 25 O2 → 18 H2O + 16 CO2
CO2 transmits visible light but absorbs heat
31
Some Interesting Chemical Reactions
3. Reduction of Iron Ore
Fe2O3 + 3 CO → 2 Fe + 3 CO2
4. Depletion of Ozone
O3 + uv rays → O2 + O
CF2Cl2 + O3 → O2 + O
32
Some Interesting Chemical Reactions
5. Photosynthesis
6 CO2 + 6 H2O → C6H12O6 + 6 O2
6. Acid Rain
S + O2 + heat → SO2
SO2 + H2O → H2SO3 (acid)
33