1. Unit 8: Stoichiometry: Part 1

2. Stoichiometry Means “element measuring”
The study of quantitative relationships between the amounts of reactants used and the products formed by a chemical reaction
Based on the law of conservation of mass

3. Mole Ratio
In a balanced equation, the ratio between the numbers of moles of any 2 substances
Indicated by the coefficients

4. 2 Mg + O2  2 MgO 2 mol Mg 1 mol O2 Mg : O2 1 mol O2 2 mol Mg 2 mol Mgor
1 mol O2
2 mol Mg
2 mol Mg
2 mol MgO
Mg : MgO or
2 mol MgO
2 mol Mg
1 mol O2
2 mol MgO
MgO : O2 or
1 mol O2
2 mol MgO

5. Stoichiometry Steps 1. Write a balanced equation.
2. Identify the given & unknown.
3. Draw a roadmap
4. Set up railroad tracks
5. Calculate

6. Mole Ratio: Use coefficients from equation
Mole  Mole
Mole to Mole Ratio
___ mol Unknown
mol Given
___ mol Given
Mole Ratio: Use coefficients from equation

7. mol O2  mol of KClO3 2KClO3  2KCl + 3O2 9 mol O2 2 mol KClO3
Ex: How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas?
2KClO3  2KCl + 3O2
? mol
9 mol
mol O2  mol of KClO3
9 mol O2
2 mol KClO3
3 mol O2
= 6 mol
KClO3

8. Mole  Mass ___ mol Unknown mol Given ___ mol Given 1 mol Unknown
Molar Mass from Periodic table
Mole to Mole Ratio
___ mol Unknown
Molar mass (g) of Unknown
mol Given
___ mol Given
1 mol Unknown
Mole Ratio: Use coefficients from equation

9. mol O2  mol of H2O  grams of H2O
Ex: How many grams of H2O would be required to produce 5 moles of O2?
2H2O  2H2 + O2
? grams
5 mol
mol O2  mol of H2O  grams of H2O
5 mol O2
2 mol
H2O
1 mol O2
18.02 g
H2O
1 mol
= g H2O

10. Mole Ratio: Use coefficients from equation
Mass  Mole
Mole to Mole Ratio
___ mol Unknown
g Given
1 mol Given
Molar mass (g) of Given
___ mol Given
Mole Ratio: Use coefficients from equation
Molar Mass
from
Periodic
table

11. grams NH3  mol NH3  mol NO 4NH3 + 5O2  4NO + 6H2O 824 g NH3 1 mol
Ex: How many moles of NO would be formed with 824 g of NH3?
4NH O2  4NO + 6H2O
824 g
? mol
grams NH3  mol NH3  mol NO
824 g
NH3
1 mol
NH3
17.04 g NH3
4 mol NO
NH3
= 48.4 mol NO

12. Mass  Mass ___ mol Unknown g Given 1 mol Given
Molar Mass from Periodic table
Mole to Mole Ratio
___ mol Unknown
Molar mass (g) of Unknown
g Given
1 mol Given
Molar mass (g) of Given
___ mol Given
1 mol Unknown
Mole Ratio: Use coefficients from equation
Molar Mass
from
Periodic
table

13. g Cu  mol Cu  mol Ag  g Ag Cu + 2AgNO3  2Ag + Cu(NO3)2 12.0 g ? g
Ex: How many grams of silver will be formed from 12.0 g copper?
Cu + 2AgNO3  2Ag + Cu(NO3)2
12.0 g
? g
g Cu  mol Cu  mol Ag  g Ag
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

14. Unit 8: Stoichiometry: Part 2

15. How many sandwiches can you make?
Available Ingredients
4 slices of bread
1 jar of peanut butter
1/2 jar of jelly
What limits the amount?
bread
What is left over?
peanut butter and jelly

16. Limiting vs. Excess Excess Reactant Limiting Reactant
The reactant that used up first in a reaction
Determines the amount of product
Excess Reactant
The reactant that is left over after the reaction stops

17. Percent Yield
The ratio of actual yield (from an experiment) to theoretical yield (from stoichiometric calculations) expressed as a percent

18. Percent Yield Actual Yield:
The amount of product actually produced when a reaction is carried out in an experiment
Theoretical Yield:
The maximum amount of product that can be produced from a given amount of reactant

19. Percent Yield Equation
measured in lab
calculated on paper

20. When 45. 8 g of K2CO3 react with excess HCl, 46. 3 g of KCl are formed
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

21. 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

22. 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%