1. Stoichiometry

2. Ham Sandwich
How many sandwiches could you make from 24 slices of bread?
How many slices of
lettuce would you need?
tomato?
ham?
cheese?

3. Equations are like recipes.
2C8H O2 → 16CO2 + 18H2O
Coefficients can be read as ratios of particles or of moles.
If 2 mols of C8H18 reacts completely how many moles of CO2 would be produced?

4. Stoichoimetry =
the proportional relationship between 2 or more substances during a chemical reaction
Predict amount of product you could make from starting amounts of reactant

5. The Mole Ratio is the Key
In stoichiometry problems, the unit that bridges the gap between one substance and another is the mole.
You can use the coefficients in conversion factors called mole ratios

6. Sample Problem A N2 + 3H2 → 2NH3
3 mols of hydrogen are needed to prepare 2 moles of ammonia. 3 mol H2 = 2 mol NH3
PROBLEM: How many moles of hydrogen are needed to prepare 312 moles of ammonia?

7. 312 mol NH3 x 3 mol H2 = ? mol H2 2 mol NH3 = 468 mol H2 needed

8. Calculate the amounts requested if 1
Calculate the amounts requested if 1.34 mol H2O2 completely react according to the following equation
2H2O2 → 2H2O + O2
mols of oxygen formed
Given: 1.34 mol H2O2 reacts
2 mol H2O2 = 1 mol O2
1.34 mol H2O2 x 1 mol O2 = mol O2 produced
2 mol H2O2

9. Calculate the amounts requested if 1
Calculate the amounts requested if 1.34 mol H2O2 completely react according to the following equation
2H2O2 → 2H2O + O2
b. mols of water formed
Given: 1.34 mol H2O2 reacts
2 mol H2O2 = 2 mol H2O
1.34 mol H2O2 x 2 mol H2O = 1.34 mol H2O produced
2 mol H2O2

10. 2. Calculate the amounts requested if 3
2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation
Fe2O3 + 2Al → 2Fe + Al2O3
mols of aluminum needed
Given: 3.30 mol Fe2O3 reacts
1 mol Fe2O3 = 2 mol Al needed
3.30 mol Fe2O3 x 2 mol Al = 6.60 mol Al needed
1 mol Fe2O3

11. 2. Calculate the amounts requested if 3
2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation Fe2O3 + 2Al → 2Fe + Al2O3 b. mols of iron formed Given: 3.30 mol Fe2O3 reacts 1 mol Fe2O3 = 2 mol Fe formed 3.30 mol Fe2O3 x 2 mol Fe = 6.60 mol Fe formed 1 mol Fe2O3

12. 2. Calculate the amounts requested if 3
2. Calculate the amounts requested if 3.30 mol Fe2O3 completely react according to the following equation Fe2O3 + 2Al → 2Fe + Al2O3 c. mols of aluminum oxide formed Given: 3.30 mol Fe2O3 reacts 1 mol Fe2O3 = 1 mol Al2O3 formed 3.30 mol Fe2O3 x 1 mol Al2O3 = 3.30 mol Al2O3 formed 1 mol Fe2O3

13. Homework
Worksheet “Stoichiometry: Mole-Mole Problems”

14. Stoichiometry: Mass-Mass Problems
You must convert to moles using molar mass of known
Then use mole ratio to find moles unknown
Convert back to mass using molar mass of unknown

15. Solving Mass-Mass Problems
Molar mass
Mass Known
Molar Ratio
Mol Known
Molar Mass
Mol Unknown
Mass Unknown
1 mol
___ grams
Periodic Table
Mol unknown
Mol known
Balanced
Chemical
Equation
___ grams
1 mol
Periodic Table

16. What mass of NH3 can be made from 1221 g H2 and excess N2
What mass of NH3 can be made from 1221 g H2 and excess N2? N2 + 3H2 → 2 NH g H2 X 1 mol H2 x 2 mol NH3 x 17.04g NH g 3 mol H2 1 mol NH3 GRAMS MOLAR MOLAR MOLAR KNOWN MASS RATIO MASS KNOWN UNKNOWN = 6867 g NH3 made

17. Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3
1. How many grams Al needed to completely react with 135 grams Fe2O3
135 g Fe2O3 X 1 mol Fe2O3 x mol Al x g Al
159.7 g Fe2O mol Fe2O mol Al
GRAMS MOLAR MOLAR MOLAR
KNOWN MASS RATIO MASS
KNOWN UNKNOWN
= 45.6 g Al needed

18. Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3
2. How many grams Al2O3 can form when 23.6g Al react with excess Fe2O3?
23.6 g Al X 1 mol Al x mol Al2O3 x g Al2O3
26.98 g Al mol Al mol Al2O3
= 44.6 g Al2O3 formed

19. Practice p. 307#1-4 Use the equation below to answer #1-4
Fe2O3 + 2Al → 2Fe + Al2O3
3. How many grams of Fe2O3 react with excess Al to make 475 g Fe?
475 g Fe X 1 mol Fe x mol Fe2O3 x g Fe2O3
55.85 g Fe mol Fe 1 mol Fe2O3
= 679 g Fe2O3 react

20. Use the equation below to answer #1-4 Fe2O3 + 2Al → 2Fe + Al2O3
4. How many grams of Fe will form when 97.6 g Al2O3 form?
97.6 g Al2O3 X 1 mol Al2O3 x mol Fe x g Fe
g Al2O mol Al2O mol Fe
= 107 g Fe formed

21. Homework
Worksheet “Stoichiometry: Mass-Mass Problems”

22. 9.2 Limiting Reactants and Percentage Yield
In this section you will:
Identify the limiting reactant for a reaction and use it to calculate theoretical yield.
Perform calculations involving percentage yield.

23. Limiting Reactant
To drive a car you need gasoline and oxygen from the air.
When the gas runs out, you can’t go any farther even though there is still plenty of oxygen.
The gasoline limits the distance you can travel because it runs out first.

24. Limiting reactant = substance that controls the quantity of product that can form in a chemical reaction; runs out first
Excess reactant = substance that is not used up completely in a reaction.
Identify the Limiting Reactant from Lab Data
Calculate the amount of product that each could form.
Whichever reactant would produce
the least amount of product is the limiting reactant.

25. Theoretical yield = maximum amount of product that can be made if everything about the reaction works out perfectly; determined by the limiting reactant.
Whenever a problem gives you quantities of 2 or more reactants, you must
Determine the limiting reactant (makes less product)
Use the limiting reactant quantity to determine the theoretical yield.

26. Identify the limiting reactant and the theoretical yield of phosphorous acid, H3PO3 if 225g of PCl3 is mixed with 123 g of H2O.
PCl3 + 3H2O → H3PO3 + 3HCl
Determine how many g of H3PO4 that each reactant could make.
Need to calculate molar masses to use as conversion factors.

27. PCl3 is the limiting reactant because 134 g is less than 187 g.
225g PCl3 x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4
137.32g mol PCl mol
= 134 g H3PO4
123g H2O x 1mol PCl3 X 1mol H3PO4 x 82.00g H3PO4
18.02g mol H2O mol
= 187 g H3PO4
PCl3 is the limiting reactant because 134 g is less than 187 g.
The theoretical yield is 134 grams of H3PO3.

28. PCl3 + 3H2O → H3PO3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair
3.00 mol PCl3 and 3.00 mol H2O
3.00 mol PCl3 x 3 mol HCl = 9.00 mol HCl
1 mol PCl3
3.00 mol H2Ox 3 mol HCl = 3.00 mol HCl
3 mol H2O
H2O is limiting reactant.
3.00 mol HCl x g = 109 grams HCl
1 mol theoretical yield (made)

29. PCl3 + 3H2O → H3PO3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair
75.0 g PCl3 and 75.0 g H2O
75.0g PCl3 x 1 mol PCl3 x 3 mol HCl = 1.65 mol HCl
136.5 g mol PCl3
75.0 g H2O x 1 mol H2O x 3 mol HCl = 4.16 mol HCl
mol H2O
PCl3 is limiting reactant.
1.65 mol HCl x g = grams HCl
1 mol theoretical yield (made)

30. PCl3 + 3H2O → H3PO3 + 3HCl Identify the limiting reactant and the theoretical yield (in grams) of HCl for each pair
mol PCl3 and 50.0 g H2O 1.00 mol PCl3 x 3 mol HCl = 3.00 mol HCl 1 mol PCl g H2O x 1 mol H2O x 3 mol HCl = 2.77 mol H3PO mol H2O H2O is limiting reactant mol HCl x g = 101 grams HCl 1 mol theoretical yield (made)

31. Limiting Reactants & Industry
The most expensive chemicals are chosen as the limiting reactants
Less expense reactants can be used in excess to ensure all of the expensive chemicals are completely used up (none wasted).

32. Actual Yield and Percentage Yield
Although equations tell you what should happen, they cannot always tell you what will happen in real life/in the lab.
Some reactions do not make all of the product predicted by the theoretical yield.
ACTUAL YIELD = the mass of product actually formed, measured in lab, often less than expected (theoretical).

33. Examples of things that could reduce yield:
Incomplete distillation/purification needed to separate product from a mixture.
Side reactions that can use up reactants without making desired products

34. PERCENTAGE YIELD = ratio relating the actual to the theoretical yield; describes how efficient the reaction was. Percentage Yield = actual x 100 theoretical SHOULD ALWAYS BE LESS THAN 100%. If not, you switched the actual & theoretical in the formula!!!