Download 22017Stoichiometry

2/17 Pre-AP
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YOUR TESTS HAVE NOT BEEN GRADED
Pick up the paper from the side table
Keep the mole packet you received yesterday.
HAVE A SEAT IN THE SAME SEAT YOU WERE ASSIGNED
YESTERDAY.
You will need a blank piece of paper and something to
write with.
We will take a large portion of the class to write a FIVE
PARAGRAPH paper over a writing prompt involving
chemistry. This is a grade and must be done in ALL core
classes. I need you to take it seriously.
I time permits, we will start stoichiometry with the mole
ratio (and the packet you received yesterday will make
more sense to you)
HW: Empirical Formula project
2/21 Pre-AP
 Pick up the papers from the side table
 HAVE A SEAT IN THE SAME SEAT YOU WERE ASSIGNED
WHEN YOU TESTED.
 WE will start stoichiometry today with mole ratio and
mole to mole calculations
 You will get your tests back on Thursday and we will
discuss the results
 4th period: you can use one of my calculators (AS LONG
AS NONE GO MISSING FROM YOUR PERIOD.
 HW: MOLE RATIO WORKSHEETS (whatever we didn’t
finish in class)
Stoichiometry
Introduction to Stoichiometry
Ideal Stoichiometric Calculations
Limiting Reactants & Percent Yield
Introduction to Stoichiometry
Reaction Stoichiometry
 Involves the mass relationships
between reactants and products
in a chemical reaction.
 4 types of problems (given, unknown)
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Mole to mole
Mole to mass
Mass to mole
Mass to mass
Conversion Factors
 There are only 2 equalities that you
will use to solve these problems!!
 Always start with a balanced
chemical equation.
 From this, you can derive the mole
ratio
 A conversion factor that relates the
amounts in moles of any two substances
involved in a chemical equation.
Mole Ratio
2Al2O3(l)  4Al(s) + 3O2(g)
 We can write mole ratios that relate any two substances
involved in the reaction.
 For example:
2 mol Al 2 O3
4 mol Al
3 mol O 2
2 mol Al 2 O3
4 mol Al
3 mol O 2
 Which substances you use and which in on top/bottom is
determined by what you are converting to/from.
Molar Mass
 The other conversion factor you may use is
the molar mass.
 Used to convert between moles and grams
of the same substance.
 Numbers are off of the periodic table and
rounded off at the hundredths position
before use.
 Ex: Molar mass of Al2O3=
 2(26.98 g/mol) + 3(16.00 g/mol)=101.96 g/mol
Ideal Stoichiometric
Calculations
4 types of problems
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Mole to mole
Mole to mass
Mass to mole
Mass to mass
General pathway:
Mass given  mole given  mole unknown  mass unknown
Mole to mole
Mass given  mole given  mole unknown  mass unknown
 Given quantity is in moles
 Unknown quantity is in moles
 Need 1 conversion factor to solve
 Mole ratio to convert between mole given &
mole unknown
# mol given mol unknown
x
 mol unknown
1
mol given
Mole to mole example
 CO2(g) + 2LiOH(s)  Li2CO3(s) + H2O(l)
How many moles of lithium hydroxide are
required to react with 20. mol of CO2?
 Given: 20. mol CO2
 Unknown: ? mol LiOH
20. mol CO2 2 mol LiOH
x
 40. mol LiOH
1
1 mol CO 2
Mole to mole example
 The elements lithium and oxygen react
explosively to form lithium oxide. How
many moles of lithium oxide will form if
2 mole of lithium react with unlimited
oxygen?
 Balanced eqn:
4Li + O2  2Li2O
 Answer: 1 mol Li2O
Mole to mass
Mass given  mole given  mole unknown  mass unknown
 Given quantity is in moles
 Unknown quantity is mass (g, kg, etc.)
 Need 2 conversion factors to solve:
 Mole ratio to convert from mole given to mole unknown
 Molar mass to convert from mole unknown to mass
unknown
# mol given mol unknown mass unknown
x
x
 mass unknown
1
mol given
1 mol unknown
Mole to mass example
 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
What mass, in grams, of glucose is
produced when 3.00 mol of water react
with carbon dioxide?
 Given: 3.00 mol water
 Unknown: ? grams glucose
3.00 mol H 2 O 1 mol C6 H12O 6 180.18 g C6 H12O 6
x
x
 90.1 g C6 H12O 6
1
6 mol H 2 O
1 mol C6 H12O 6
Mole to mass example
 2NaN3(s)  2Na(s) + 3N2(g)
If 0.500 mol of NaN3 react, what mass
in grams of N2 would result?
 Given: 0.500 mol NaN3
 Unknown: ? grams N2
 Answer: 21.0 g N2
Mass to mole
Mass given  mole given  mole unknown  mass unknown
 Given quantity is a mass (g, kg, etc.)
 Unknown quantity is in moles
 Need 2 conversion factors to solve:
 Molar mass to convert from mass given to mole given
 Mole ratio to convert from mole given to mole unknown
# g given 1 mol given mol unknown
x
x
 mol unknown
1
g given
mol given
Mass to mole example
 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
This reaction is run using 824 g NH3 &
excess O2. How many moles of NO are
formed?
 Given: 824 g NH3
 Unknown: ? mol NO
824 g NH3 1 mol NH3
4 mol NO
x
x
 48.4 mol NO
1
17.04 g NH3 4 mol NH3
Mass to mole example
 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
This reaction is run using 824 g NH3 &
excess O2. How many moles of H2O
are formed?
 Given: 824 g NH3
 Unknown: ? mol H2O
 Answer: 72.6 mol H2O
AP CHEM….
 Take out a clean piece of paper
 Answer the following WITHOUT looking at
Kinetics notes
 1. Define Kinetics.
 2. What needs to happen to reactant molecules in
order for successful collisions to occur to increase
the rate of a reaction?
 3. List and EXPLAIN three factors that affect the
rate of a reaction.
 4. How much of a temperature increase will
double the reaction rate?
 5. Describe how a catalyst such as manganese
oxide speeds up a reaction.
AP
 Pick up the graded papers from the
side table
 Leave your IMF test and Ch8/Ch9
Quiz out
 Turn in the Practice AP Test
3/17 WELCOME BACK!
 Pick up your graded papers or wait until
someone passes them out.
 On the back of the stoichiometry quiz,
answer the following:
1. 4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
This reaction is run using 824 g NH3 &
excess O2. How many grams of NO are
formed?
2. 2NaN3(s)  2Na(s) + 3N2(g)
If 0.500 mol of NaN3 react, how many
moles of N2 would result?
4NH3(g) + 5O2(g)  4NO(g) + 6H2O(g)
This reaction is run using 824 g NH3 & excess
O2. How many grams of NO are formed?
2NaN3(s)  2Na(s) + 3N2(g)
If 0.500 mol of NaN3 react, how many moles of
N2 would result?
Today you will need…
 A calculator, a periodic table, the mass to
mass worksheet you picked up Friday
 Take out the mole ratio WS and TURN IN IF
COMPLETED
 If you didn’t finish, you have ten minutes
 Answer the following on a piece of paper:
 The elements lithium and oxygen react
explosively to form lithium oxide. How many
moles of lithium oxide will form if 2 mole of
lithium react with unlimited oxygen?
 The elements lithium and oxygen react
explosively to form lithium oxide. How
many moles of lithium oxide will form if
2 mole of lithium react with unlimited
oxygen?
Mass to mass
Mass given  mole given  mole unknown  mass unknown
 Given quantity is a mass (g, kg, etc.)
 Unknown quantity is a mass
 Need 3 conversion factors to solve:
 Molar mass to convert from mass given to mole given
 Mole ratio to convert from mole given to mole
unknown
 Molar mass to covert from mole unknown to mass
unknown
# g given 1 mol given mol unknown mass unknown
x
x
x
 mass unknown
1
g given
mol given
1 mol unknown
Mass to mass example
 NH4NO3(s)  N2O(g) + 2H2O(l)
How many grams of NH4NO3 are
required to produce 33.0 g N2O?
 Given: 33.0 g N2O
 Unknown: ? g NH4NO3
33.0 g N 2 O 1 mol N 2 O 1 mol NH 4 NO3 80.06 g NH 4 NO3
x
x
x
 60.0 g NH 4 NO3
1
44.02 g N 2 O
1 mol N 2 O
1 mol NH 4 NO3
AP
 Take out Kinetics notes
Mass to mass example
 What mass of aluminum is produced
by the decomposition of 5.0 kg of
Al2O3?
 Given: 5.0 kg Al2O3
 Unknown: ? g Al
 Balanced eqn:
2Al2O3  4Al + 3O2
 Answer: 2.6 kg
Today you will need…
 The lab paper from the side table
 Pick up YOUR graded papers
 A calculator, a periodic table, something to
write with
 Take out the mass to mass homework
 We will practice mole to mole and mass to
mass problems today
 We will also prepare for the lab you have
Wednesday/Thursday
Add the following questions to your
mass to mass HW from last night.
These will count as 8, 9, 10
In a very violent reaction called a thermite reaction,
aluminum metals reacts with iron(III) oxide to form
iron metal and aluminum oxide according to the
following equation: Fe2O3 + Al  Fe + Al2O3.
8. What is the mole ratio of Al to Fe? Of aluminum
oxide to iron (III) oxide?
9. What mass of Fe will be produced from 130. g
of Fe2O3?
10. How many moles of Al will react with 2.58
moles of Fe2O3?
Ideal conditions
 These problems tell us the amount of
reactants or products under ideal
conditions.
 All reactants are completely converted
into products.
 Give the maximum yield we could
expect, but this is rarely attained in the
field because we don’t have ideal
conditions.
4th/6th
 Need:
 Paper from table, calculator, something
to write with
 On a clean piece of paper:
 Explain limiting reactant
 Explain excess reactant
(this will be turned in)
Limiting Reactants & Percent
Yield
Limiting & excess reactants
 Once one of the reactants is completely
used up in a rxn, it doesn’t matter how
much of the other reactant(s) you have.
The reaction cannot continue.
 Limiting reactant:
 The reactant that limits the amounts of the
other reactants that can combine and the
amount of product that can form in a
chemical reaction.
Limiting & excess reactants
 Excess reactant(s)
 Substance(s) that are not completely
used up an a rxn and do not limit the
amount of product that can be formed.
 These are the reactants that are “left
over” at the end of a rxn.
Which is limiting? Excess?
 To decide which reactant is limiting in
a rxn, use one of your givens to solve
for the other.
 Then compare how much you have
(given) to how much you need under
ideal conditions (solved for).
 If you have more than you need, the 1st
substance is limiting.
 If you have less than you need, the 2nd
substance is limiting.
Limiting reactant example
 CO(g) + 2H2(g)  CH3OH
If 500. mol of CO and 750. mol of H2 are
present, which is the limiting reactant?
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Solve to determine how much H2 would be needed to
completely react 500. mol CO.
500. mol CO 2 mol H 2
x
 10 0 0 mol H 2
1
1 mol CO
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Do you have 1000 mol H2? No, you only have 750. mol.
H2 is the limiting reactant.
Limiting reactant example
3ZnCO3(s) + 2C6H8O7(aq)  Zn3(C6H5O7)2(aq) + 3H2O(l) + 3CO2(g)
 If there is 1 mol of ZnCO3 & 1 mol of C6H8O7,
which is the limiting reactant?
 Answer:
 1 mol ZnCO3 could react with 0.67 mol C6H8O7,
which is less than is available.
 ZnCO3 is limiting.
Limiting reactant example
 Aspirin, C9H8O4, is synthesized by the rxn of
salicylic acid, C7H6O3, with acetic anhydride,
C4H6O3.
2C7H6O3 + C4H6O3  2C9H8O4 + H2O
 When 20.0g of C7H6O3 and 20.0g of C4H6O3 react,
which is the limiting reactant? How many moles of the
excess reactant are used when the rxn is complete?
What mass in grams of aspirin is formed?
 C7H6O3, 0.0724 mol, 26.1 g
Percent Yield
 Theoretical yield-maximum amount of
product that can be produces from a
given amount of reactant.
 This is what we calculate using ideal
stoichiometric calculations.
 Actual yield-the measured amount of
a product obtained from a rxn.
Percent Yield
 Ratio of the actual yield to the
theoretical yield, multiplied by 100.
actual yield
% yield 
x 100
theoretica l yield
Percent Yield example
 C6H6(l) + Cl2(g)  C6H5Cl(s) + HCl(g)
When 36.8g C6H6 react with an excess of Cl2, the
actual yield of C6H5Cl is 38.8g. What is the percent
yield?
 Given:
36.8g C6H6, excess Cl2, actual yield=38.8g C6H5Cl
 Unknown: ? g C6H5Cl, percent yield
36.8 g C 6 H 6
1 mol N 2 O 1 mol C 6 H 5Cl 112.56 g C 6 H 5Cl
x
x
x
 53.0 g C6 H 5Cl
1
78.12 g C 6 H 6 1 mol C6 H 6
1 mol C 6 H 5Cl
% yield 
38.8 g C6 H 5Cl
actual yield
x 100 
x 100  73.2%
theoretica l yield
53.0 g C6 H 5Cl
Percent Yield example
 Methanol can be produced through the
rxn of CO and H2 in the presence of a
catalyst.
CO(g)  2H2(g)  CH3OH(l)
catalyst
 If 75.0 g of CO reacts to produce 68.4
g CH3OH, what is the percent yield?
 Answer: 79.7%
Percent Yield example
 2ZnS(s) + 3O2(g)  2ZnO(s) + 2SO2(g)
If the typical yield is 86.78%, how much SO2
should be expected if 4897 g of ZnS are
used?
Theoretica l yield 
4897 g ZnS 1 mol ZnS 2 mol SO 2 64.07 g SO 2
x
x
x
 3219.278 g SO 2
1
97.46 g ZnS 2 mol ZnS 1 mol SO 2
actual yield
(% yield)(the oretical yield)
% yield 
x 100  actual yield 
theoretica l yield
100
(86.78)(32 19.278 g SO 2 )
actual yield 
 2794 g SO 2
100