the study of quantitative, or measurable,...
TRANSCRIPT
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Chapter 9
Stoichiometry
• The study of quantitative, or
measurable, relationships that
exist in chemical formulas and
chemical reactions.
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Mole – Mole Relationship
• Use a balanced chemical equation to
determine the mole – mole
relationship of any substance.
• Mole – Mole relationship can be
applied to any substance in the
balanced equation.
• N2H
4 + 2 H
2O
2 N
2 + 4 H
2O
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Practice Mole-Mole
• Determine the number mole-
mole relationship for Ag in the
following equation:
• 3 AgNO3 + Al Al(NO
3)3 + 3 Ag
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Mole – Mole Problems
• How many moles of HCl are
needed to react with 2.3 moles
of Zn?
• 2 HCl + Zn ZnCl2 + H
2
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Practice Problems
• 1) How many moles of Na are
needed to react with 55.5 moles
of Cl2,
in the production of salt?
Na + Cl2 NaCl
• 2) How many moles of Al(OH)3 are
produced when 2.53 moles of H2O
react.
Al2O
3 + H
2O Al(OH)
3
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Answers
1) 2 Na + Cl2 2NaCl
2) Al2O
3 + 3 H
2O 2 Al(OH)
3
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9-2 Ideal Stoichiometric
Calculations
• Using all other conversions from
chapter 7, along with the mole-
mole to determine unknown
amounts of each substance.
• 1 mol = 6.02 x 1023
particles
• 1 mol = 22.4L at STP
• 1 mol = formula mass(g)
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Example: Mass - Mass
• Determine the mass of sodium
hydroxide produced when .25g of
sodium reacts with water.
2 Na + 2H2O 2NaOH + H
2
.25g x g
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Practice Problem #1
• What mass of bromine, is
produced when fluorine reacts
with 3.55 moles of potassium
bromide?
F2 + 2 KBr 2 KF + Br
2
3.55 mol x g
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Mass – Volume :
Example
• Find the mass of aluminum
required to produce 1.32 L of
hydrogen gas at STP.
2 Al + 3 H2SO
4 Al
2(SO
4)3 + 3 H
2
x g 1.32 L
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Volume – Volume:
Relationship
• The volume relationship of 2
gases at STP are directly
proportional to the mole-mole
relationship of those 2 gases.
• Why?
• All gases at STP occupy the
same volume, 22.4L / mol.
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Example:
Volume - Volume
• What volume of hydrogen gas is
necessary to produce 16.0L of
ammonia (NH3)?
N2 + 3 H
2 2 NH
3
x L 16.0 L
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Chapter 9–3: Limiting
and Excess Reactants
• Limiting reactant
– The reactant that limits the amount
of product obtained.
– Completely consumed in the
reaction. No trace remains in the
product.
• Excess reactant
– The reactant that isn’t completely
used in the reaction and remains
after the reaction is over.
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Example Problem:
• Determine the mass of NaCl produced
when 1.2 mol of Na react with .85 mol
Cl2.
2 Na + Cl2 2 NaCl
1.2mol .85mol ?mol
1st determine the limiting reactant.
Convert moles of one given to moles of
another.
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Determining Limiting
and Excess
• Compare: Have vs. Need
• Have = Given amount
• Need = Calculated amount
• If Have > Need, excess reactant
• If Need > Have, limiting reactant
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Comparing
• Determine the mass of NaCl
produced when 1.2 mol of Na react
with .85 mol Cl2.
2 Na + Cl2 2 NaCl
1.2mol .85mol ?g
22 Cl .6mol
Na 2mol
Cl 1mol
1
molNa2.1x
You have .85mol Cl2 you need .6mol
Cl2.
Chlorine is the excess, which means
Sodium is the limiting.
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Finishing the Problem
• Determine the mass of NaCl produced
when 1.2 mol of Na react with .85 mol
Cl2.
2 Na + Cl2 2 NaCl
1.2mol .85mol ?g
You must begin with the limiting given
to determine the unknown amount.
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Practice Problem
3ZnO + 2Al(NO3)3 Al
2O
3 + 3Zn(NO
3)2
Considering 25g of ZnO react with
.555moles of Al(NO3)3. How many
grams of Al2O
3 will be produced?
a) Determine limiting/excess reactant.
b) Determine excess that will remain.
c) Determine the unknown.
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• A)
.
• B)
• C)
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Percent Yield
• Yield = Product
• Actual Yield
– Amount of product given in the
problem.
• Theoretical Yield (Expected)
– Amount of product determined
by a stoichiometry calculation.
• Percent Yield
100 x yield ltheoretica
yield actual YieldPercent
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Sample Problem
• Determine the percent yield
when 2.80g Al(NO3)3 react to
produce .966g of Al(OH)3.
Al(NO3)3 + 3NaOH Al(OH)
3 + 3NaNO
3
2.80g .966g
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Identifying what to do!
• 1) Regular stoichiometry:
• Given one value.
• 2) Limiting/Excess:
• Given 2 values both are reactants (left
side of equation)
• 3) %yield:
• Given 2 values 1 reactant and 1 product
• All problems involve a mole-mole
conversion.
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Practice Problem
• Determine the percent yield when
1.50g Al(OH)3 decomposes to produce
.25g of H2O.
2Al(OH)3 Al
2O
3 + 3H
2O
1.50g .25g