Transcript PPT File

Agenda 1/28/2013
• Turn in work that is due or had been
missing
• Standard check – 3f and 3c (3b, 3d)
• Limiting reactants and percent yield
(mandatory listening/write 2 definitions)
• “A” Student Assignment
• The Mole is a number…(mandatory notes)
• Computer lab /Project peer review, spelling,
formatting check (MLA)
• Homework – complete “World’s …est”
Article. Due 11:30 pm Weds turnitin.com
Limiting Reactants (or why do
reactions stop?)
wood +
oxygen in air
→water vapor
+ carbon
dioxide
What is the limiting reactant?
Why use an excess of a
reactant?
Why not use exact mole ratios
given in balanced equation?
Some reactions stop before
reach completion (probability of
collisions)
inefficient/wasteful
Why use an excess of a
reactant? Cont.
Over time scientists have found
often more efficient to use an
excess of one reactant
(least expensive)
“drives” reaction until all limiting
reactant used up
Can also speed up a reaction
Why use an excess of a
reactant?
Limited air (oxygen)
yellow – glowing bits
of unburned fuel
(sooty)
Correct
fuel/oxygen ratio
– complete
combustion
Stoichiometry
This is the slide that must be copied.
Section 12.3 Limiting Reactants
Use your text to define each term.
New
Vocabulary
limiting
reactant
excess
reactant
164
Stoichiometry
This is the slide that must be copied.
Section 12.3 Limiting Reactants
Use your text to define each term.
New
Vocabulary
limiting
reactant
excess
reactant
164
Limits the extent of the chemical
reaction and thereby determines
the amount of product
Stoichiometry
This is the slide that must be copied.
Section 12.3 Limiting Reactants
Use your text to define each term.
New
Vocabulary
limiting
reactant
excess
reactant
164
Limits the extent of the chemical
reaction and thereby determines
the amount of product
“leftover” or unused reactants in a
chemical reaction
Percent Yield
Fe2O3 + 3CO →2Fe + 3CO2
• 1 mole (160g) of iron (III) oxide will produce
2 moles of pure iron (2 x 56)g or 112g
• 112g is the “theoretical yield” (maximum
amount of product possible
• In practice what happens?
• Get less than expected, say 110g “actual
yield”
Percent Yield
Fe2O3 + 3CO →2Fe + 3CO2
Percent yield = actual yield (from experiment) x 100
Theoretical yield(from stoichiometric calculation)
• 112g is the “theoretical yield” for Fe
• 110g “actual yield” of Fe
Percent yield Fe = 110g x 100 = 98%
112g
“A” Student Assignment – Deadline
Friday, February 8th
Read section 12.3 Limiting Reactants
Complete notes page 165, 1st 2 sub-headings only (stop
when you get to Determining the Limiting Reactant).
Read section 12.4 Percent Yield
Complete notes pages 167, 168, and 169
On lined paper complete Section 12.3 Assessment
Questions 22 through 26 and Section 12.4 Assessment
Questions 30 through 34. Create suitable titles.
To turn this work in the three Science notebook pages and
the separate papers all need to be stapled together
before class. Name need only be written on first page.
Carbon-12
isotope is
stable and
readily
available
all over
the world
The Mole is the SI base unit for the
amount of a substance.
The quantity one mole is set by
defining one mole of carbon-12
atoms to have a mass of exactly 12
grams.
(CST 3b)
The Mole as a Number of Particles
This we
already
should
know
Standard 3c
“representative
particles”
The mole is the SI unit to measure
the amount of a substance. 12g of
carbon-12 is set to define one mole
and contains one mole of carbon-12
atoms.
One mole equals
6.02 x 1023 particles of the
substance (atoms,
molecules, ions).
The Mole is a Unit song
http://www.youtube.com/watch?v=1R7NiIum2TI&NR=1
Think about how we assign a name to numbers
all the time:
A dozen eggs/roses/bread rolls
2 dozen
5 dozen
million = 1 000 000
=1.0 x 106
23
10
The Mole is 6.02 x
particles (Avogadro’s constant)
Fe2O3 + 3CO → 2Fe + 3CO2
1 mole
3 moles
2 moles
3 moles
6.02 x 1023 (formula units) Fe2O3
3 x (6.02 x 1023) = 18.0 x 1023
= 1.80 x 1024
molecules of CO
The Mole is 6.02 x
particles
23
10
Fe2O3 + 3CO → 2Fe + 3CO2
1 mole
3 moles
2 moles
3 moles
2 x (6.02 x 1023) atoms Fe
12.0 x 1023 = 1.2 x 1024
3 x (6.02 x 1023) = 18.0 x 1023
= 1.80 x 1024
molecules of CO2
Scientific Notation – Appendix B p 889-891
http://www.nyu.edu/pages/mathmol/textbook/scinot.html
CST Question – When methane gas is burned in
the presence of oxygen, the following chemical
reaction occurs
Combustion & Exothermic reaction
CH4(g) + 2O2(g) 
CO2(g) + 2H2O(g)
If 1 mole of methane reacts with 2
moles of oxygen, then
(and the answers are in …. Molecules of CO2 and H2O.
What do you do?
CST Questions – convert from number of moles
to number of molecules by multiplying the
number of moles by NA (Avogadro’s constant)
CH4(g) + 2 O2(g) 
1 mole 2 moles
CO2(g) + 2 H2O(g)
1 mole
6.02x1023
6.02x1023
Molecules CO2
2 moles
2x(6.02x1023)
1.20x1024
molecules H2O
Get used to it/comfortable with it
1 mole
2 moles
3 moles
6.02 x 1023 particles
2x(6.02x1023) = 1.20x1024
3(6.02 x 1023) = 1.8 x 1024
10 moles 10(6.02 x 1023) = 6.02 x 1024
0.5 moles 0.5(6.02 x 1023) = 3.01 x 1023
0.25 moles 0.25(6.02 x 1023) = 1.5 x 1023
0.1 moles 0.1(6.02 x 1023) =6.02 x 1022
Avogadro’s Principle: equal volumes of gases
at the same temperature and pressure
contain equal numbers of particles.
22.4L
22.4L
He
H2
Cl2
28.2 cm
28.2 cm
22.4L
22.4L
Molar Volume: A mole (6.02 x 1023 particles)
of any gas occupies 22.4 L at STP (0°C which
is 273 K and 1 atm).
Planning
Mon
Jan
Feb
Tue
s
Weds
22
30 3d and
ionic names
Project
4 ionic names &
review
6Test – Ch 6,
7, all mole,
equations,
naming cmpds
hol
Fri
24 (3e)
Mass/equ
ations
28 mole/no. of
particles 3c,d
Computer lab
time (peer
review)
hol
Thurs
12
8 Lab
5wk
Grades
14
20
22
lab