1. Stoichiometry

2. 9.1 Calculating Quantities in Reactions
Determine mole ratios from a balanced chemical equation
Explain why mole ratios are central to solving stoichiometry problems
Solve stoichiometry problems involving:
Mass using molar mass
Volume using denisty
Particles using Avogadro’s number

3. Ham Sandwich
How many sandwiches could you make from 24 slices of bread?
How many slices of
lettuce would you need?
tomato?
ham?
cheese?
This process models the calculations in this chapter.

4. 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?

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

6. 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

7. 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?

8. 312 mol NH3 x 3 mol H2 = ? mol H2 2 mol NH3 = 468 mol H2 needed

9. Practice Prob. P. 304
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

10. Practice Prob. P. 304
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

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 Thermite Reaction
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

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 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

13. 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

14. Application/Homework
Worksheet “Stoichiometry: Mole-Mole Problems” – front only

15. 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

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

17. Sample Problem B p. 307
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

18. Practice p. 307#1-4 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

19. Practice p. 307#1-4 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

20. 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

21. Practice p. 307#1-4 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

22. Homework
Worksheet “Stoichiometry: Mass-Mass Problems” – back of mol-mol worksheet Lab: Mass Relations in Chemical Reactions (Baking soda & HCl reaction)

23. Stoichiometry: Volume-Volume Problems
Liquid amounts often measured in volumes. You might use density and molar mass for these problems p. 308 Toolkit
Density unit g/mL means ____g = 1 mL
Gases L = 1 mol of any gas
Basics the same: Change to moles, use mole ratio, and then change to desired units.

24. Solving Volume-Volume Problems
Volume Unknown
Volume Known
Mass Known
Molar mass
Mol Known
Molar Ratio
Mol Unknown
Molar Mass
Density
Density
Mass Unknown

25. Sample Problem C p. 309
What volume of H3PO4 forms when 56 mL POCl3 completely react? (density of POCl3 = 1.67 g/mL; density of H3PO4 = 1.83 g/mL)
POCl3(l) + 3H2O(l) → H3PO4 (l) + 3HCl (g)
56mL POCl3 x 1.67 g POCl3 x 1 mol POCl x 1 mol H3PO4
1 mL g mol POCl3
X g H3PO4 x 1 mL H3PO = mL H3PO4
1 mol g

26. Practice p. 309 # 1-4
C5H12 (l) → C5H8(l) + 2H2(g) Densities: C5H12 = g/mL C5H8 = g/mL H2 = g/L = g/1000mL = 8.99 x 10-5 g/mL Calculate Molar Masses: C5H12 = g/mol C5H8 = g/mol H2 = 2.02 g/mol

27. How many mL of C5H8 can be made from 366mL C5H12?
366mL C5H12 x g x 1 mol C5H12 x 1 mol C5H8
1 mL g mol C5H12
X g x 1 mL
1 mol C5H g
= 315 mL C5H8

28. 2. How many liters of H2 can form when 4.53 x 103 mL C5H8 form?
4.53 x 103 mLC5H8 x g x 1 mol C5H8 x 2 mol H2
1 mL g mol C5H8
X g x 1 L
1 mol H g
= 2030 L H2

29. 3. How many mL of C5H12 are needed to make 97.3 mL of C5H8?
97.3 mL C5H8 x g x 1 mol C5H8 x 1 mol C5H12
1 mL g mol C5H8
X g x 1 mL
1 mol C5H g
= 113 mL C5H12

30. 4. How many milliliters of H2 can be made from 1.98 x 103 mL C5H12?
1.98 x 103 mL C5H12 x g x 1 mol C5H12 x 2 mol H2
1 mL g mol C5H12
X g x 1 mL
1 mol H g
= 7.64 x 105 mL H2

31. Homework WS “Stoichiometry: Volume-Volume Problems” front only
WS “Stoichiometry: MixedProblems” back – classwork
9.1 QUIZ will be on __________________

32. 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.

33. Limiting Reactant Page 312/313 Figures 3&4
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.
Page 312/313 Figures 3&4
Which of the starting supplies limited the number of mums that could be made? Which items where in excess/left over?

34. 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.

35. 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.

36. Determine how many g of H3PO4 that each reactant could make.
Sample E p. 314
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.

37. 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.

38. Practice p. 314 #1-3 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)

39. Practice p. 314 #1-3 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)

40. Practice p. 314 #1-3 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)

41. 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).

42. 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).

43. 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

44. 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!!!

45. Sample Problem F p. 317 N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.
14.0g N2 x 1 mol N2 x 2 mol NH3 x g NH3 = 17.0g NH3
28.02 g mol N mol NH3
9 .0g H2 x 1 mol N2 x 2 mol NH3 x g NH3 = 51 g NH3
2.02 g mol H mol NH3
N2 makes the smaller amount and is the limiting reactant.

46. Sample Problem F p. 317 N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.
N2 makes the smaller amount and is the limiting reactant.
The theoretical amount made is 17.0g NH3.
The actual amount made is 16.1 g NH3
PERCENTAGE YIELD = 16.1 g x 100 = 94.7%
17.0g

47. Practice p. 317 #1
N2 + 3 H2 → 2NH3
1. Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 , 3.15g H2 and actual is 14.5 g NH3.
14.0g N2 x 1 mol N2 x 2 mol NH3 x g NH3 = 17.0g NH3
28.02 g mol N mol NH3
3.15g H2 x 1 mol N2 x 2 mol NH3 x g NH3 = 17.7 g NH3
2.02 g mol H mol NH3
N2 makes the smaller amount and is the limiting reactant.

48. N2 + 3 H2 → 2NH3
Determine the limiting reactant, theoretical yield and percentage yield if 14.0g N2 are mixed with 9.0 g H2 and 16.1 g NH3 form.
N2 makes the smaller amount and is the limiting reactant.
The theoretical amount made is 17.0g NH3.
The actual amount made is 14.5 g NH3
PERCENTAGE YIELD = 14.5 g x 100 = 85.3%
17.0g

49. Homework
Section 9.2 Review p. 319 # 1,3,4,5,6,8,10,12,14 Quiz on 9.2 will be on __________________.