1. Lesson 3
Percentage Yield
and Energy

2. PERCENTAGE YIELD AND PERCENTAGE PURITY:
Percentage Yield: is used to describe the amount of product actually obtained as a percentage of the expected amount.
2 reasons:
1. reactants may not all react
2. some products are lost
Yield: is the amount made in a chemical reaction

3. THREE TYPES of YIELD 
Actual Yield – what you get in the lab when the chemicals are mixed
Theoretical Yield – what the balanced equation says you should make
Percent Yield Actual Yield x 100%
Theoretical Yield

4. Sometimes reactions do not go to completion
Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%.
1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%?
2Fe2O3 + 3C  4Fe + 3CO2
100. g ? g
100. g Fe2O3

5. Sometimes reactions do not go to completion
Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%.
1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%?
2Fe2O3 + 3C  4Fe + 3CO2
100. g ? g
100. g Fe2O3 x 1 mole
159.6 g

6. Sometimes reactions do not go to completion
Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%.
1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%?
2Fe2O3 + 3C  4Fe + 3CO2
100. g ? g
100. g Fe2O3 x 1 mole x 4 mole Fe
159.6 g 2 mole Fe2O3

7. Sometimes reactions do not go to completion
Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%.
1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%?
2Fe2O3 + 3C  4Fe + 3CO2
100. g ? g
100. g Fe2O3 x 1 mole x 4 mole Fe x g
159.6 g 2 mole Fe2O mole

8. Sometimes reactions do not go to completion
Sometimes reactions do not go to completion. Reaction can have yields from 1% to 100%.
1. How many grams of Fe are produced by the reaction of 100. g of Fe2O3, if the percentage yield is 75.0%?
2Fe2O3 + 3C  4Fe + 3CO2
100. g ? g
100. g Fe2O3 x 1 mole x 4 mole Fe x g x = g
159.6 g 2 mole Fe2O mole

9. Percentage Yield = Actual Yield x 100%
Theorectical Yield
Actual Yield is what is experimentally measured.
Theoretical Yield is what is calculated using stoichiometry.

11. 2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of
2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.
75.1 g actual yield
2AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq)
152 g ? g
152. g AgNO3 x 1 mole
169.9 g

12. 2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of
2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.
75.1 g actual yield
2AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq)
152 g ? g
152. g AgNO3 x 1 mole x 1 mole Ag2SO4
169.9 g mole AgNO3

13. 2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of
2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.
75.1 g actual yield
2AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq)
152 g ? g
152. g AgNO3 x 1 mole x 1 mole Ag2SO4 x g = g
169.9 g mole AgNO mole

14. 2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of
2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.
75.1 g actual yield
2AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq)
152 g ? g
152. g AgNO3 x 1 mole x 1 Ag2SO x g = g
169.9 g mole AgNO mole
% yield = x %

15. 2. In an experiment 152. g of AgNO3 is used to make 75. 1 g of
2. In an experiment 152. g of AgNO3 is used to make 75.1 g of Ag2SO4(s). Calculate the percentage yield.
75.1 g actual yield
2AgNO3(aq) + Na2SO4(aq)  Ag2SO4(s) + 2NaNO3(aq)
152 g ? g
152. g AgNO3 x 1 mole x 1 Ag2SO x g = g
169.9 g mole AgNO mole
% yield = x % = 53.8 %
139.5

16. Energy Calculations
The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in an endothermic or exothermic reaction.
3. How much energy is required to produce 25.4 g of H2?
213 kJ H2O  2H2 + O2
? kJ g

17. Energy Calculations
The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in an endothermic or exothermic reaction.
3. How much energy is required to produce 25.4 g of H2?
213 kJ H2O  2H2 + O2
? kJ g
25.4 g H2

18. Energy Calculations
The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in an endothermic or exothermic reaction.
3. How much energy is required to produce 25.4 g of H2?
213 kJ H2O  2H2 + O2
? kJ g
25.4 g H2 x 1 mole
2.02 g

19. Energy Calculations
The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in an endothermic or exothermic reaction.
3. How much energy is required to produce 25.4 g of H2?
213 kJ H2O  2H2 + O2
? kJ g
25.4 g H2 x 1 mole x kJ
2.02 g 2 mole H2

20. Energy Calculations
The energy term in a balanced equation can be used to calculate the amount of energy consumed or produced in an endothermic or exothermic reaction.
3. How much energy is required to produce 25.4 g of H2?
213 kJ H2O  2H2 + O2
? kJ g
25.4 g H2 x 1 mole x kJ = x 103 kJ
2.02 g 2 mole H2

21. 4. How many molecules of H2 can be produced when 452 kJ of
4. How many molecules of H2 can be produced when 452 kJ of energy if consumed?
2H2 + O2  2H2O kJ
? Molecules kJ
452 kJ

22. 4. How many molecules of H2 can be produced when 452 kJ of
4. How many molecules of H2 can be produced when 452 kJ of energy if consumed?
2H2 + O2  2H2O kJ
? Molecules kJ
452 kJ x 2 moles H2
213 kJ

23. 4. How many molecules of H2 can be produced when 452 kJ of
4. How many molecules of H2 can be produced when 452 kJ of energy if consumed?
2H2 + O2  2H2O kJ
? Molecules kJ
452 kJ x 2 moles H2 x x molecules
213 kJ mole

24. 4. How many molecules of H2 can be produced when 452 kJ of
4. How many molecules of H2 can be produced when 452 kJ of energy if consumed?
2H2 + O2  2H2O kJ
? Molecules kJ
452 kJ x 2 moles H2 x x molecules = x molecs
213 kJ mole

25. 4. How many molecules of H2 can be produced when 452 kJ of
4. How many molecules of H2 can be produced when 452 kJ of energy if consumed?
2H2 + O2  2H2O kJ
? Molecules kJ
452 kJ x 2 moles H2 x x molecules = x molecs
213 kJ mole

26. 5. How much energy is produced by an explosion of a 5. 2 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2  2H2O kJ
5.2 L ? kJ

27. 5. How much energy is produced by an explosion of a 5. 2 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2  2H2O kJ
5.2 L ? kJ
5.2 L

28. 5. How much energy is produced by an explosion of a 5. 2 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2  2H2O kJ
5.2 L ? kJ
5.2 L x 1 mole
22.4 L

29. 5. How much energy is produced by an explosion of a 5. 2 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2  2H2O kJ
5.2 L ? kJ
5.2 L x 1 mole x kJ
22.4 L moles H2

30. 5. How much energy is produced by an explosion of a 5. 2 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2  2H2O kJ
5.2 L ? kJ
5.2 L x 1 mole x kJ = kJ
22.4 L moles H2
Home work
Worksheet # 3 page 131

31. Problems with Molarity
Often it is necessary to determine how much water to add to a solution to change it to a specific concentration.
The concentration of a solution is typically given in molarity. Molarity is defined as the number of moles of solute (what you are dissolving) divided by the liters of solvent (what is being dissolved into).
Molarity (M)=#mole/Litre of solution

32. A student wants 50.0 L of hydrogen gas at STP in a plastic bag by reacting excess aluminum metal with 3.00 M (moles per litre) sodium hydroxide solution according to the reaction…
2NaOH + 2Al +6H2O → 2NaAl(OH)4 + 3H2
What volume of sodium hydroxide solution is required?

33. 2NaOH + 2Al +6H2O → 2NaAl(OH)4 + 3H2
3.00 M L
? Vol.
What volume of sodium hydroxide solution is required?
Note: use Molarity as your mole to volume conversion

34. HCl (aq) + NaOH H2O(l) + NaCl(l) ? vol 25.0mL
What volume of M HCl is required to completely neutralize 25.0mL of 0.318M NaOH?
HCl (aq) + NaOH H2O(l) + NaCl(l)
? vol mL
0.250 M M
Note: you are using the Molarity (moles/L) as your moles and volume conversion!

36. Remember:

37. How to determine an unknown Molarity: Ex 1:
1. #moles KOH = M x vol
= 0.500mol x 0.025L L
= mol

38. How to experimentaly determine an Unknown Molarity:
Titration
v2w
digital-lab-techniques-manual-spring- 2007/videos/titration/

39. Limiting/Excess Reactants
Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the limiting reactant.
Sometimes a reaction will occur between two or more substances where one substance is in excess. This substance is referred to as being an excess reactant.
Often, it is necessary to identify the limiting or excess reactant in a problem.

40. What mass of Br2 is produced when 25. 0 grams of K2CrO7, 55
What mass of Br2 is produced when grams of K2CrO7, 55.0 grams of KBr and grams of H2SO4 are reacted according to the equation below? How many grams of each excess reactant will remain unreacted?
K2CrO7 + 6KBr +7H2SO4 4K2SO4 + Cr2(SO4)3 +3Br2 + 7H2O

41. Mass of Product
Based on

43. Masses based on KBr

44. Masses in Excess