Download Stoichiometry notes 1

Stoichiometry – Ch. 9
I.
Stoichiometric
Calculations
(p. 275-287)
A. Proportional Relationships
2 1/4 c. flour
1 tsp. baking soda
1 tsp. salt
1 c. butter
3/4 c. sugar
3/4 c. brown sugar
1 tsp vanilla extract
2 eggs
2 c. chocolate chips
Makes 5 dozen cookies.
I
have 5 eggs. How many cookies
can I make?
Ratio of eggs to cookies
5 eggs 5 doz.
2 eggs
= 12.5 dozen cookies
A. Proportional Relationships
 Stoichiometry
• mass relationships between
substances in a chemical reaction
• based on the mole ratio
 Mole
Ratio
• indicated by coefficients in a
balanced equation
2 Mg + O2  2 MgO
B. Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature
&
0°C and 1 atm
Pressure
Note: When not at STP the ideal gas law will be used to
convert between moles ↔ liters of gas (cover this later)
C. Avogadro’s Number
Number of particles in one mole of a substance.
1 mole = 6.022 x 1023 particles
Can be used as a conversion factor between moles and
Number of representative particles.
Reminder:
The representative particles for
1. an element is an atom
2. for a molecular compound is a molecule
3. for an ionic compound is a formula unit
D. Molarity
Molarity is a unit of concentration – it tells us the amount
of solute in one liter of solution
Molarity can be used to convert between moles and
volume of solution.
Molarity, M = moles of solute / liters of solution
or
Molarity, M = moles of solute / 1000 ml of solution
E. Stoichiometry Steps
1. Write a balanced equation.
2. Label given and target.
3. Convert given unit to moles using the appropriate
conversion factors.
Conversion Factors:
moles ↔ grams
moles ↔ number of particles
moles ↔ liters of solution
moles ↔ liters of gas at STP
Molar Mass
Avogadro’s Number
Molarity
Standard Molar Volume
4. Convert moles of given substance to moles of target
substance using the mole ration from the balanced
equation.
Note: This is the core step in all stoichiometry problems!
5. Convert moles of target to final unit using the
appropriate conversion factor.
C. Molar Volume at STP
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS
IN
GRAMS
Molar Mass
(g/mol)
6.02 
MOLES
1023
particles/mol
Molarity (mol/L)
LITERS
OF
SOLUTION
NUMBER
OF
PARTICLES
D. Stoichiometry Problems
 How
many moles of KClO3 must
decompose in order to produce 9
moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol O2 2 mol KClO3
3 mol O2
9 mol
= 6 mol
KClO3
D. Stoichiometry Problems
many grams of KClO3 are req’d to
produce 9.00 L of O2 at STP?
 How
2KClO3  2KCl + 3O2
?g
9.00 L
9.00 L
O2
1 mol
O2
2 mol 122.55
KClO3 g KClO3
22.4 L
O2
3 mol
O2
1 mol
KClO3
= 32.8 g
KClO3
D. Stoichiometry Problems
 How
many grams of silver will be
formed from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
?g
12.0 1 mol 2 mol 107.87
g Cu Cu
Ag
g Ag
= 40.7 g
63.55 1 mol 1 mol
Ag
g Cu
Cu
Ag
D. Stoichiometry Problems
 How
many grams of Cu are required
to react with 1.5 L of 0.10M AgNO3?
Cu + 2AgNO3  2Ag + Cu(NO3)2
?g
1.5L
0.10M
1.5 .10 mol 1 mol 63.55
L AgNO3
Cu
g Cu
1L
= 4.8 g
2 mol 1 mol
Cu
AgNO3 Cu
Stoichiometry – Ch. 9
II. Stoichiometry in the
Real World
(p. 288-294)
A. Limiting Reactants
Available
Ingredients
• 4 slices of bread
• 1 jar of peanut butter
• 1/2 jar of jelly
Limiting
Reactant
• bread
Excess
Reactants
• peanut butter and jelly
A. Limiting Reactants
Limiting
Reactant
• used up in a reaction
• determines the amount of
product
Excess
Reactant
• added to ensure that the other
reactant is completely used up
• cheaper & easier to recycle
A. Limiting Reactants
1. Can be identified by the question
involving amounts of more than one
reactant.
2. For each reactant, calculate the
amount of product formed using the
5 steps of stoichiometry.
3. Smaller answer indicates:
• limiting reactant
• amount of product
A. Limiting Reactants
Reminder - The five steps of stoichiometry
1. Write a balanced chemical equation.
2. Label your given and target substances.
3. Convert your given unit(s) to moles of given substance
using the appropriate conversion factor.
4. Convert moles of given substance to moles of target
substance using the mole ratio from the balanced
equation.
5. Convert moles of target substance to the final unit using
the appropriate conversion factor.
Note: Carry out steps 3-5 as needed for both reactants in
a limiting reactant problem.
A. Limiting Reactants
 79.1
g of zinc react with 0.90 L of
2.5M HCl. Identify the limiting and
excess reactants. How many liters
of hydrogen are formed at STP?
Zn + 2HCl 
79.1 g 0.90 L
2.5M
ZnCl2 + H2
?L
A. Limiting Reactants
Zn + 2HCl 
79.1 g 0.90 L
2.5M
ZnCl2 + H2
?L
79.1 1 mol 1 mol 22.4 L
g Zn Zn
H2
H2
= 27.1 L
65.39 1 mol 1 mol
H2
g Zn
Zn
H2
A. Limiting Reactants
Zn + 2HCl 
79.1 g 0.90 L
2.5M
0.90 2.5 mol 1 mol
L
HCl
H2
1L
ZnCl2 + H2
?L
22.4
L H2
= 25 L
2 mol 1 mol
H2
HCl
H2
A. Limiting Reactants
Zn: 27.1 L H2
HCl: 25 L H2
Limiting reactant: HCl
Excess reactant: Zn
Product Formed: 25 L H2
left over zinc
B. Percent Yield
measured in lab
actual yield
% yield 
 100
theoretical yield
calculated on paper
B. Percent Yield
When
45.8 g of K2CO3 react
with excess HCl, 46.3 g of KCl
are formed. Calculate the
theoretical and % yields of KCl.
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
B. Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
?g
actual: 46.3 g
Theoretical Yield:
45.8 g 1 mol 2 mol 74.55
K2CO3 K2CO3
KCl g KCl
= 49.4
138.21 g 1 mol 1 mol g KCl
K2CO3 K2CO3 KCl
B. Percent Yield
K2CO3 + 2HCl  2KCl + H2O + CO2
45.8 g
49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
% Yield =
46.3 g
49.4 g
 100 = 93.7%