1. Unit 8 Stoichiometry Notes
Balancing Equations, Stoichiometry, Limiting Reactants, % Yield

2. Stoichiometry Island Diagram
Known Unknown
Substance A Substance B M
Mass
Mass Mountain
Mass
Mole Island
Volume
Mole
Mole
Volume V
Liter Lagoon
Particles
Particles P
Particle Place
Stoichiometry Island Diagram

3. Stoichiometry Island Diagram
Known Unknown
Substance A Substance B
Mass
Mass
1 mole = molar mass (g)
1 mole = molar mass (g)
Use coefficients
from balanced
chemical equation
Volume
Mole
Mole
Volume
1 mole = STP
1 mole = STP
(gases)
(gases)
1 mole = x 1023 particles
(atoms or molecules)
1 mole = x 1023 particles
(atoms or molecules)
Particles
Particles
Stoichiometry Island Diagram

4. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10 10
___ mole Cl2 5 5
___ mole NaCl 10 10 ?

5. Visualizing a Chemical Reaction2
Na Cl NaCl 2
___ mole Na 10
___ mole Cl2 5
___ mole NaCl 10

6. Formation of Ammonia N2 (g) + 3 H2 (g) 2 NH3 (g) + + + + + + + +
2 atoms N and
2 atoms N +
6 atoms H
6 atoms H
1 molecule N2 +
3 molecules H2
2 molecules NH3
10 molecule N2 +
30 molecules H2
20 molecules NH3
6.02 x 1023
molecules N2
6.02 x 1023
molecules H2
6.02 x 1023
molecules NH3
1 x +
3 x
2 x
1 mol N2 +
3 mol H2
2 mol NH3
28 g N2 +
3 x 2 g H2
2 x 17 g NH3
34 g reactants
34 g products
Assume
STP
22.4 L
22.4 L
22.4 L
22.4 L
22.4 L
22.4 L +
22.4 L N2
67.2 L H2
44.8 L NH3

7. Standard Temperature & Pressure
Molar Volume at STP
1 mol of a gas=22.4 L
at STP
Standard Temperature & Pressure
0°C and 1 atm
Courtesy Christy Johannesson

8. Molar Volume at STP MOLES LITERS OF GAS AT STP (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
Courtesy Christy Johannesson

9. Mole  Mole Stoichiometric Calculation
(GIVEN AMOUNT)
(Mole Ratio from BALANCED Chemical Reaction)
Given moles
Unknown mole ratio
Known mole ratio
2 : : 2
Given 3 moles of O2. How many moles of H2O produced?
2H2 + O2  2H2O
3 moles O2
2 moles H2O
1 mole O2
= 6 moles H2O

10. Mole  Mass Stoichiometric Calculation
(GIVEN AMOUNT)
(Molar Mass)
(Mole Ratio from BALANCED Chemical Reaction)
Given mass
1 mol known
Unknown mole ratio
mass known
Known mole ratio
2 : : 2
Given 4 moles of H2. How many grams of H2O produced?
2H2 + O2  2H2O
4 moles H2
2 moles H2O
18.02 g H2O
2 mole H2
1 mol H2O
= grams H2O

11. Mass  Mass Stoichiometric Calculation
(GIVEN AMOUNT)
(Molar Mass Known)
(Mole Ratio from BALANCED Chemical Reaction)
(Molar Mass Unknown)
Given mass
1 mol known
Unknown mole ratio
mass unknown
mass known
Known mole ratio
1 mol unknown
2 : : 2
How many grams of H2 are needed to react with 64 grams of O2?
2H2 + O2  2H2O
64g O2
1 mole O2
2 mol H2
2.02g H2
32g O2
1 mol O2
1 mol H2
= 8.08 grams H2

12. Mole  Liters Stoichiometric Calculation
(GIVEN AMOUNT)
(Mole Ratio from BALANCED Chemical Reaction)
Liter/mol conversion
Given moles
Unknown mole ratio
22.4 L unknown
Known mole ratio
1 mol unkown
2 : : 2
Given 3 moles of O2. How many liters of H2 needed?
2H2 + O2  2H2O
3 moles O2
2 moles H2
22.4 L H2
1 mole O2
1 mol H2
= Liters H2

13. Stoichiometry Practice Problems
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol KClO3
Courtesy Christy Johannesson

14. 2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3
How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl O2
? mol
9 mol
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol KClO3
2 mol KClO3
6 mol
x mol KClO3 = 9 mol O2
= 6 mol KClO3
3 mol O2 O2
KClO3
Courtesy Christy Johannesson

15. Stoichiometry Problems
How many grams of KClO3 are required to produce 9.00 L of O2 at STP?
2KClO3  2KCl O2
? 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
= g
KClO3
Courtesy Christy Johannesson

16. Stoichiometry Problems
How many grams of KClO3 are required to produce 9.00 L of O2 at STP?
2KClO3  2KCl O2
? g
9.00 L
1 mol O2
2 mol KClO3
g KClO3
x g KClO3 = 9.00 L O2
= 32.8 g KClO3
22.4 L O2
3 mol O2
1 mol KClO3
32.8 g O2
KClO3
Courtesy Christy Johannesson

17. How many grams of KClO3 are required to produce 9.00 L of O2 at STP?
2KClO3  2KCl O2
? 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
= g
KClO3
32.8 g
1 mol O2
2 mol KClO3
g KClO3
x g KClO3 = 9.00 L O2
= 32.8 g KClO3
22.4 L O2
3 mol O2
1 mol KClO3 O2
KClO3
Courtesy Christy Johannesson

18. How many grams of silver will be formed from 12.0 g copper?
Cu AgNO3  2 Ag Cu(NO3)2
? g
12.0 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
40.7 g
1 mol Cu
2 mol Ag
g Ag
x g Ag = 12.0 g Cu
= 40.7 g Ag
63.55 g Cu
1 mol Cu
1 mol Ag Cu Ag
Courtesy Christy Johannesson

19. Calculating Limiting Reactants
STEPS:
Balance chemical reaction
Choose a reactant and calculate the amount of product.
Choose the other reactant and calculate the amount of product.
Whichever reactant makes LESS product, that is your limiting reactant.
The LEAST amount of product produced from Steps 3 and 4, is the amount theoretically produced by the reaction.