Transcript Slide 1

Stoichiometry
• APPLICATION:
• We are applying concepts learned in
Chapters 5, 6 & 7
» Naming/Formulas
» Molar Conversions
» Balancing Chemical Equations
• PURPOSE:
• To understand how chemical equations
are used to express ratios of
reactants and products
1
Stoichiometry
• Stoichiometry:
– From Greek:
– “Stoichio” = matter or chemicals
– “Metry” = measure
=The calculation of quantities
in a chemical reaction.
• Kinds of quantities measured:
•
•
•
•
Particles (atoms, molecules, formula units)
Moles
Mass
Volume
2
Stoichiometry
Why is it useful?
• To determine the mass, volume or number of
particles needed for a particular reaction.
• To determine the mass, volume or number of
particles produced by a particular reaction.
Who uses stoichiometry?
•
•
•
•
•
•
Chemists,
Chemical engineers,
Engineers,
Rocket scientists,
Environmental scientists….
+ criminals making illegal drugs, terrorists making bombs…
3
Stoichiometry
• Why are balanced chemical equations
useful?
• Look at the ratios of all reactants and products
10
1 N2(g)
+ 30
3 H2(g)
2
20 NH3 (g)
• One molecule of nitrogen reacts with three
molecules of hydrogen to produce two
molecules of ammonia
» 1:3:2 ratio
• What happens if you have 10 molecules of N2?
4
5
Stoichiometry
• Mass:
• All balanced chemical reactions must
follow the Law of Conservation of Mass
1
N2(g) +
3
H2(g)
• 1 mol N2 = 28.0 g
• 3 mol H2 = 6.0 g
• 2 mol NH3 = 34.0 g
2
NH3 (g)
Reactants = 34.0 g
Products = 34.0 g
6
Stoichiometry
• Volume:
• At STP (standard temperature and
pressure); 1 mol of any gas = 22.4 L
1
N2(g) +
3
H2(g)
2
NH3 (g)
• Is volume conserved?
Reactants = 89.6 L
• 1 mol N2 = 22.4 L
• 3 mol H2 = 67.2 L (3 x 22.4)
Products = 44.8 L
• 2 mol NH3 = 44.8 L (2 x 22.4)
VOLUME IS NOT USUALLY CONSERVED!
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Stoichiometry
• Moles:
• The coefficients in a balanced chemical
equation represent the number of moles
of that substance
1
N2(g) +
3
H2(g)
2
NH3 (g)
• Are # of Moles conserved?
• 1 mol N2
• 3 mol H2
Reactants = 4 mol
• 2 mol NH3
Products = 2 mol
MOLES ARE NOT USUALLY CONSERVED!
8
Stoichiometry
• Moles:
• All coefficients are related to the
number of moles required for a
reaction.
1
N2(g) +
3
H2(g)
2
NH3 (g)
• One mole of nitrogen reacts with
three moles of hydrogen to produce
two moles of ammonia
KEY CONCEPT: Coefficients show
relative numbers of moles involved
in a reaction = the molar ratio
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Stoichiometry
• Mole-Mole Calculations
1
N2(g) +
3
H2(g)
2
NH3 (g)
• Balanced chemical equations are valuable
because they shoe the relative ratio of
reactants to products.
• You can now relate the # of moles of a
reactant(s) to find the # of moles of a
product(s)
10
Stoichiometry
• Sample Problem 1:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– How many moles of ammonia are
produced when 1.2 moles of nitrogen
reacts with hydrogen?
11
Stoichiometry
• Example:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– Step 1:
• Look at balanced chemical equation
» 1 mol of N2 produces 2 mol NH3
– Step 2:
• Use a conversion factor
» Convert mol N2 to mol NH3
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Stoichiometry
• Example: How much NH3 can be made
from 1.2 mol N2?
1
N2(g) +
3
1.2 mol N2
H2(g)
2 mol NH3
1 mol N2
2
NH3 (g)
= 2.4 mol NH3
Mole ratio
13
Stoichiometry
• Sample Problem 2:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– Calculate the number of moles of nitrogen
required to make 9.54 mol of NH3.
9.54 mol NH3
1 mol N2
2 mol NH3
= 4.77 mol N2
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Stoichiometry
• Sample Problem 3:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– Calculate the number of moles of hydrogen
required to make 9.54 mol of NH3.
9.54 mol NH3
3 mol H2
2 mol NH3
= 14.3 mol H2
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16
Stoichiometry
• Mass-Mass Calculations:
– When scientist conduct a lab experiment,
they measure the amount of a particular
substance using grams…not moles
– Remember, you can convert the mass of a
substance into moles (1-step conversion)
Mass (grams) A  moles A
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Stoichiometry
• Mass-Mass Calculations:
– Additionally, it is possible to convert
between masses of reactants and products.
– The mole ratio (obtained from the
balanced chemical equation) is the key to
convert from the mass of substance A to
mass of substance B
mass A  moles A  moles B  mass B
– Remember, the balanced chemical equation
provides the mole ratio!
18
Stoichiometry
• Mass-Mass Example #1:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– Calculate the number of grams of NH3
produced by the reaction of 6.80 g of
hydrogen with an excess of nitrogen. Use
the balanced equation above.
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Stoichiometry
• Mass-Mass Example #1:
1
N2(g) +
3
H2(g)
2
NH3 (g)
– Step 1: What are we calculating?
• Grams H2  grams NH3
– Step 2: What is the path we need to take?
mass H2  moles H2  moles NH3  mass NH3
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Stoichiometry
• Mass-Mass Example #1:
1
N2(g) +
3
2
H2(g)
NH3 (g)
– Step 3: Complete conversions
6.80 g H2
1 mol H2
2.0 g H2
2 mol NH3
17.0 g NH3
3 mol H2
1 mol NH3
= 38.5 g NH3
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Stoichiometry
• Mass-Mass Example #2:
1
N2(g) +
3
2
H2(g)
NH3 (g)
• How many grams of nitrogen are needed to
produce the 38.5 g of NH3 produced in the
previous example?
38.5 g NH3
1 mol NH3
17.0 g NH3
1 mol N2
28.0 g N2
2 mol NH3
1 mol N2
= 31.7 g N2
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Stoichiometry
• Mass-Mass Example #3:
• What is the total amount of mass (in grams) of
the reactants in the example balanced chemical
equation?
Is mass conserved in a
chemical reaction???
Mass of Reactants = Mass of Products
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Stoichiometry
• Mass-Mass Example #3:
Mass of Reactants = Mass of Products
31.7 g N2 + ___ g H2  38.5 g NH3
ANSWER  38.5 – 31.7 = 6.8 g H2
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Stoichiometry
• Other Stoichiometric Calculations:
– You can use the same steps as before to
calculate the following…
» Mass-Volume
» Volume-Volume
» Particle-Mass
– The first step is to ALWAYS convert to
moles
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Stoichiometry
• Other Stoichiometric Calculations:
– Choose the appropriate road map…
particles A
mass A
volume A
particles B
moles A
moles B
mass B
volume B
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Limiting Reagent
• Limiting Reagent
• Chemical equations are like recipes…
» The reactants are combined to make the products
• Reactions take place based upon the mole ratios
expressed in the balanced chemical equation
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Limiting Reagent
• Limiting Reagent
• Limiting Reagent:
– Limits how much product will form =
determines the amount of product that can
be formed in a reaction
• Excess Reagent:
– The substance(s) that is leftover because
they are in excess = there is more than
enough to react with the limiting reagent
A reaction can only occur until the
limiting reagent is used up!!
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Limiting Reagent
• Limiting Reagent  S’mores
2
•
•
•
•
Graham Cracker = Gc
Marshmallow = M
Chocolate = Ch
S’more = Gc2MCh
Gc +
1
M +
1
Ch
1
Gc2MCh
To make one s’more…you need to have 2
graham crackers, 1 marshmallow, and 1
piece of chocolate.
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• Limiting Reagent  S’mores
2
Gc +
1
M +
1
Ch
1
Gc2MCh
• How many s’mores can you make if you have
14 (mol) graham crackers, 10 pieces of
chocolate, and 8 marshmallows?
14 mol Gc
1 mol Gc2MCh
2 mol Gc
10 mol Ch
1 mol Gc2MCh
1 mol Ch
8 mol M
1 mol Gc2MCh
1 mol M
= 7.0 mol Gc2MCh
= 10.0 mol Gc2MCh
= 8 mol Gc2MCh
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• Limiting Reagent  S’mores
2
Gc +
1
M +
1
Ch
1
Gc2MCh
• How many s’mores can you make if you have
14 (mol) graham crackers, 10 pieces of
chocolate, and 8 marshmallows?
• Graham crackers  limiting reagent
• THE LIMITING AGENT WILL DETERMINE
HOW MUCH PRODUCT WILL BE CREATED!
14 mol Gc
1 mol Gc2MCh
2 mol Gc
= 7.0 mol Gc2MCh
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Limiting Reagent
• Limiting Reagent
1
N2(g) +
3
H2(g)
2
NH3 (g)
• What would happen if 2 mol of N2 reacted with
3 mol of H2?
• NEED: Nitrogen-Hydrogen ratio  1:3
• HAVE: Nitrogen-Hydrogen ratio  2:3
• What reactant is limiting?
• What reactant is in excess?
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Limiting Reagent
• Limiting Reagent
1
N2(g) +
2 mol N2
3
H2(g)
3 mol H2
1 mol N2
3 mol H2
1 mol N2
3 mol H2
2
NH3 (g)
6 mol H2
needed
But don’t have!
1 mol N2
needed
have!
There is not enough hydrogen for 2 mol N2,
so: hydrogen is the limiting reagent!!
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•
2
Limiting Reagent—Example #1
Na(s) +
•
1
Cl2(g)
2
NaCl (s)
What will occur when 7.20 mol of Na reacts
with 3.5 mol of Cl2?
a) What is the limiting reagent?
b) How many moles of NaCl are produced?
c) How much of the excess reagent remains
unreacted?
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•
2
Limiting Reagent—Example #1
Na(s) +
•
1
Cl2(g)
2
NaCl (s)
What will occur when 7.20 mol of Na reacts
with 3.5 mol of Cl2?
a) Choose one of the reactants and convert to
the amount of moles of the other reactant.
7.20 mol Na
1 mol Cl2
2 mol Na
= 3.6 mol Cl2
Does the value you calculated exceed the
amount presented in the problem???
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•
2
Limiting Reagent—Example #1
1
Na(s) +
•
Cl2(g)
2
NaCl (s)
What will occur when 7.20 mol of Na reacts
with 3.5 mol of Cl2?
b) Use the given amount (from the problem) of
the limiting reagent…calculate moles of NaCl.
moles Cl2  moles NaCl
3.5 mol Cl2
2 mol NaCl
1 mol Cl2
= 7.0 mol NaCl
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•
2
Limiting Reagent—Example #1
Na(s) +
•
1
Cl2(g)
2
NaCl (s)
What will occur when 7.20 mol of Na reacts
with 3.5 mol of Cl2?
c) The amount of excess reagent is the
difference between the given amount of Na
and the amount of Na needed in the reaction.
moles Cl2 moles Na
3.5 mol Cl2
2 mol Na
1 mol Cl2
= 7.0 mol Na
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•
2
Limiting Reagent—Example #1
Na(s) +
•
1
Cl2(g)
2
NaCl (s)
What will occur when 7.20 mol of Na reacts
with 3.5 mol of Cl2?
c) The amount of excess reagent is the
difference between the given amount of Na
and the amount of Na needed in the reaction.
7.20 mol Na – 7.00 mol Na = 0.20 mol Na in excess
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•
2
Limiting Reagent—Example #2
Cu(s) +
•
1
S (g)
1
Cu2S (s)
What is the maximum number of grams of
Cu2S that can be formed when 80.0 g of Cu
reacts with 25.0 g of S?
a) Find the number of moles of each reactant.
80.0 g Cu
1 mol Cu
= 1.26 mol Cu
63.5 g Cu
25.0 g S
1 mol Cu
32.1 g S
= 0.779 mol S
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•
2
Limiting Reagent—Example #2
Cu(s) +
•
1
S (g)
2
Cu2S (s)
What is the maximum number of grams of
Cu2S that can be formed when 80.0 g of Cu
reacts with 25.0 g of S?
b) Use the balanced chemical equation to
determine the limiting reagent.
1.26 mol Cu
1 mol S
= 0.630 mol S
2 mol Cu
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•
2
Limiting Reagent—Example #2
Cu(s) +
•
1
S (g)
2
Cu2S (s)
Use the balanced chemical equation to
determine the limiting reagent.
Compare the amount of sulfur calculated
to the amount of sulfur presented in the
problem.
0.630 mol S  calculated
0.779 mol S  problem
= 0.630 mol S
Sulfur is in
excess…copper is
limiting
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•
2
Limiting Reagent—Example #2
Cu(s) +
•
1
2
S (g)
Cu2S (s)
What is the maximum number of grams of
Cu2S that can be formed when 80.0 g of Cu
reacts with 25.0 g of S?
c) Use the amount of limiting reagent to
calculate the maximum amount of the product.
1.26 mol Cu
1 mol Cu2S
159.1 g Cu2S
2 mol Cu
1 mol Cu2S
= 100.2 g Cu2S
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Percent Yield
• Percent Yield:
• This is a comparison of how much of a given
product is created in a reaction to the ideal
yield of the product in an ideal reaction.
• Theoretical Yield:
» Using an equation to determine the amount of
product formed during a reaction
• Actual Yield:
» The amount of a product created in an actual
laboratory experiment
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Percent Yield
• Percent Yield:
Percent Yield
=
Actual Yield
Theoretical Yield
X 100%
• Percent yield will NEVER exceed 100%
• Percent yield can be less than 100%
• Comparison of MASS; not moles
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• Percent Yield  Example #1
• What is the percent yield of this reaction if
34.8 g of CaCO3 is heated to give 16.4 g of CaO?
1
CaCo3 (s)
1
CaO(s) +
1
CO2 (g)
• Determine the actual yield of CaO from the
problem
» 16.4 g CaO
• Calculate the theoretical yield of CaO.
mass CaCO3  moles CaCO3  moles CaO  mass CaO
34.8 g CaCO3
1 mol CaO
1 mol CaO
56.1 g CaO
100.1 g CaCO3 1 mol CaCO3 1 mol CaO
= 19.5 g CaO48
• Percent Yield  Example #1
• What is the percent yield of this reaction if
34.8 g of CaCO3 is heated to give 16.4 g of CaO?
1
CaCo3 (s)
Percent Yield =
Percent Yield =
1
CaO(s) +
Actual Yield
Theoretical Yield
16.4 g CaO
19.5 g CaO
1
CO2 (g)
X 100
X 100
= 84.1 %
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