3/18/2014

Today:

Schedule an appointment or stop
by office hours to discuss QUIZ 2

Next Meeting
◦ Reading for Wednesday:
◦ Stoichiometric Analysis:
 Mole to Mole Conversions:
 Use stoichiometric coefficients
from balanced equations
 Gram to Gram Conversions:
 Start Chapter 6, pp. 214-225
 Use MOLAR MASS to get to moles
 Theoretical Yield:
 How much product can form in a
reaction?
 Limiting Reagents:
 Method 1
 Method 2
iClicker Participation Question:
Hybridization & Orbital Diagrams of Molecules
What is the hybridization of the HIGHLIGHTED carbon atom?
A. sp
B. s2p
C. sp2
D. sp3
E. sp4
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3/18/2014
Stoichiometry:
Studying the quantities in chemical reactions
Law of Conservation of Mass: requires the same number
of atoms on each side of the chemical equation.

Helpful in predicting the amount of
products that can form based on the
amount of starting reactants
2 NaN3(s)

2 Na(s) + 3 N2(g)
2 sodium azide units will decompose
into just 2 sodium atoms & 3
nitrogen molecules.
Sodium azide decomposition is
used to inflate air bags.
Mole-to-Mole Conversions:
Stoichiometric coefficients express PARTICLE ratios
2 NaN3(s)
2 Na(s) + 3 N2(g)
How many moles of N2 could form from 6 moles of NaN3?
Theoretical Yield: MAXIMUM amount of product that can form
in a reaction based on the limited starting materials.
How many moles of NaN3 are needed to produce 2.67 moles of N2 gas
(about the amount of gas needed to fill one airbag)?
iClicker Participation Question:
Mole-to-Mole Conversions—From Reactants to Products
How much NH3 could be produced, if 40 moles of hydrogen
gas reacted according to the equation below?
A. 80 moles
N2(g) + 3 H2(g)
2 NH3(g)
B. 60 moles
C. 40 moles
D. 27 moles
E. 20 moles
“The expansion of the world's population
from 1.6 billion people in 1900 to today's
seven billion [in 2012] would not have been
possible without the [industrial] synthesis
of ammonia.“ –MIT Press
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3/18/2014
Mass-to-Mass Conversions:
How many grams of sodium metal can be produced from the
decomposition of 130 grams of sodium azide?
# of grams
NaN3
2 NaN3(s)
# of grams
Na
2 Na(s) + 3 N2(g)
Moles cannot be measured directly, but…
Molar Mass
of NaN3:
22.99 +
3 x 14.01 =
65.02 g/mol
MASS can be measured on a balance &
tied to the number of moles with
MOLAR MASS
Mole-to-mole ratios relate
reactants & products:
# of moles
NaN3
2 NaN3 produce 2 Na
Molar Mass
Na:
22.99 g/mol
# of moles
Na
What volume of water could form from the
combustion of 20. mL of C3H8O?
2 C3H8O + 9 O2(g)
6 CO2(g) + 8 H2O(g)
2 mol of C3H8O
make 8 mol H2O
# of moles
C3H8O
MOLAR MASS = 60.1 g/mol
# of moles
H2O
MOLAR MASS = 18.0 g/mol
Mass of
C3H8O
Mass of
H2O
DENSITY = 0.786 g/mL
DENSITY = 1.00 g/mL
Volume of
C3H8O
Volume of
H2O
Limiting Reagents:
2 C3H8O + 9 O2(g)
Trial 1:
• For the reaction to occur, C3H8O
must combine with O2 in a 2 to
9 ratio
• 20 mL C3H8O reacts O2 in bottle
to produce CO2, H2O, & HEAT
6 CO2(g) + 8 H2O(g)
Trial 2:
• Without
NO REACTION
2, there isReagent:
O2 is theOLimiting
It
• LIMITS
Inside the
O2 isofdepleted
thebottle:
amount
product
• Reaction is LIMITED to the
that can form. The other
opening where O2 is present
reactant is in EXCESS.
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3/18/2014
How many grams of oxygen would be needed to
completely react with 20. mL of C3H8O?
2 C3H8O
# of moles
C3H8O
+
9 O2(g)
2 mol of C3H8O
react with 9 mol O2
MOLAR MASS = 60.1 g/mol
# of moles
O2
MOLAR MASS = 32.0 g/mol
Mass of
C3H8O
Mass of
O2
DENSITY = 0.786 g/mL With LESS than 38 g O : the 20 mL of
2
C3H8O cannot fully react. This would
Volume of
make the O2 the limiting reagent &
C3H8O
the C3H8O would be in excess.
Stoichiometric Coefficients:
A recipe for making (chemical) products
• The stoichiometric coefficients specify the exact PARTICLE ratio
necessary to convert the reactants to the products.
4H
4
+
+ 1
1C
1 CH4
1
It will always require 4 wheels and 1 frame to make the core structure of a car:
• 3 wheels & 1 frame could not make a complete product. In this scenario,
the wheels would be LIMITING the production of a complete car.
• 120 wheels & 20 frames: more frames would be necessary to use up all
the available wheels. The frames would be LIMITING. Without more
frames, some unused wheels would be left over.
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