1. Stoichiometric Calculations (p. 275-287)
Stoichiometry – Ch. 9
Stoichiometric Calculations (p )

2. 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

3. 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

4. Standard Temperature & Pressure
B. Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm
Note: When not at STP the ideal gas law will be used to
convert between moles ↔ liters of gas (cover this later)

5. C. Avogadro’s Number Number of particles in one mole of a substance.
1 mole = x 1023 particles
Can be used as a conversion factor between moles and
Number of representative particles.
Reminder:
The representative particles for
an element is an atom
for a molecular compound is a molecule
for an ionic compound is a formula unit

6. 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

7. E. Molar Mass Molar Mass serves as the conversion factor between
grams and moles. It has the units of grams/1 mole
grams → moles given grams x 1 mole / ? Grams
moles → grams given moles x ? Grams / 1 mole
? Stands for gram-formula or gram – atomic mass of
the substance.

8. F. Stoichiometry Steps Write a balanced equation.
Label given and target.
Convert given unit to moles using the appropriate
conversion factors.
Conversion Factors:
moles ↔ grams Molar Mass
moles ↔ number of particles Avogadro’s Number
moles ↔ liters of solution Molarity
moles ↔ liters of gas at STP 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!
Convert moles of target to final unit using the
appropriate conversion factor.

9. G. Connections between units
LITERS
OF GAS
AT STP
Molar Volume
(22.4 L/mol)
MASS IN
GRAMS
MOLES
NUMBER OF
PARTICLES
Molar Mass
(g/mol)
6.02  1023
particles/mol
Molarity (mol/L)
LITERS OF
SOLUTION

10. H. Stoichiometry Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol
KClO3

11. Stoichiometry Problems
How many grams of KClO3 are req’d to produce 9.00 L of O2 at STP?
2KClO3  2KCl + 3O2
? g
9.00 L
9.00 L O2
1 mol O2
22.4 L
2 mol
KClO3
3 mol O2
122.55
g KClO3
1 mol
KClO3
= 32.8 g
KClO3

12. 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
g Cu
1 mol Cu
63.55
g Cu
2 mol Ag
1 mol Cu
107.87
g Ag
1 mol Ag
= 40.7 g Ag

13. 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 L
.10 mol
AgNO3
1 L
1 mol Cu
2 mol
AgNO3
63.55
g Cu
1 mol Cu
= 4.8 g Cu

14. II. Stoichiometry in the Real World (p. 288-294)
Stoichiometry – Ch. 9
II. Stoichiometry in the Real World (p )

15. A. Limiting Reactants § Limiting Reactant bread § Excess Reactants
§ Available Ingredients
4 slices of bread
1 jar of peanut butter (50 oz)
1/2 jar of jelly (24 oz)
§ Recipe – 2 slices of bread, 1 oz jelly, 2 oz
peanut butter.
§ Limiting Reactant
bread
§ Excess Reactants
peanut butter and jelly

16. 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

17. 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

18. A. Limiting Reactants Reminder - The five steps of stoichiometry
Write a balanced chemical equation.
Label your given and target substances.
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.
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.

19. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g ? L 0.90 L 2.5M
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 HCl  ZnCl H2
79.1 g
0.90 L
2.5M
? L

20. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L
g Zn
1 mol Zn
65.39
g Zn
1 mol H2 Zn
22.4 L H2
1 mol
= 27.1 L H2

21. A. Limiting Reactants Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L 2.5M ? L
2.5 mol
HCl
1 L
1 mol H2
2 mol
HCl
22.4
L H2
1 mol H2
= 25 L H2

22. 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

23. B. Percent Yield
measured in lab
calculated on paper

24. 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

25. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g
actual: 46.3 g
Theoretical Yield:
45.8 g
K2CO3
1 mol
K2CO3 g
2 mol
KCl
1 mol
K2CO3
74.55
g KCl
1 mol
KCl
= 49.4
g KCl

26. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g 49.4 g
actual: 46.3 g
Theoretical Yield = 49.4 g KCl
46.3 g
49.4 g
% Yield =
 100 =
93.7%