lesson 3 percentage yield and energy. sometimes reactions do not go to completion. reaction can have...
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Lesson 3Percentage Yield
and Energy
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
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
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
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 55.8 g
159.6 g 2 mole Fe2O3 1 mole
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 55.8 g x 0.750 = 52.4 g
159.6 g 2 mole Fe2O3 1 mole
Percentage Yield = Actual Yield x 100%Theorectical Yield
Actual Yield is what is experimentally measured. Theoretical Yield is what is calculated using stoichiometry.
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
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 Ag2SO4
169.9 g 2 mole AgNO3
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 Ag2SO4 x 311.9 g = 139.5 g
169.9 g 2 mole AgNO3 1 mole
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 Ag2SO4 x 311.9 g = 139.5 g
169.9 g 2 mole AgNO3 1 mole
% yield = 75.1 x 100 %
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 Ag2SO4 x 311.9 g = 139.5 g
169.9 g 2 mole AgNO3 1 mole
% yield = 75.1 x 100 % = 53.8 % 139.5
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 + 2H2O 2H2 + O2
? kJ 25.4 g
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 + 2H2O 2H2 + O2
? kJ 25.4 g
25.4 g H2
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 + 2H2O 2H2 + O2
? kJ 25.4 g
25.4 g H2 x 1 mole2.02 g
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 + 2H2O 2H2 + O2
? kJ 25.4 g
25.4 g H2 x 1 mole x 213 kJ2.02 g 2 mole H2
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 + 2H2O 2H2 + O2
? kJ 25.4 g
25.4 g H2 x 1 mole x 213 kJ = 1.34 x 103 kJ2.02 g 2 mole H2
4. How many molecules of H2 can be produced when 452 kJ of
energy if consumed?
2H2 + O2 2H2O + 213 kJ
? Molecules 452 kJ
452 kJ
4. How many molecules of H2 can be produced when 452 kJ of
energy if consumed?
2H2 + O2 2H2O + 213 kJ
? Molecules 452 kJ
452 kJ x 2 moles H2
213 kJ
4. How many molecules of H2 can be produced when 452 kJ of
energy if consumed?
2H2 + O2 2H2O + 213 kJ
? Molecules 452 kJ
452 kJ x 2 moles H2 x 6.02 x 1023 molecules 213 kJ 1 mole
4. How many molecules of H2 can be produced when 452 kJ of
energy if consumed?
2H2 + O2 2H2O + 213 kJ
? Molecules 452 kJ
452 kJ x 2 moles H2 x 6.02 x 1023 molecules = 2.55 x 1024 molecs 213 kJ 1 mole
4. How many molecules of H2 can be produced when 452 kJ of
energy if consumed?
2H2 + O2 2H2O + 213 kJ
? Molecules 452 kJ
452 kJ x 2 moles H2 x 6.02 x 1023 molecules = 2.55 x 1024 molecs 213 kJ 1 mole
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2 2H2O + 213 kJ
5.2 L ? kJ
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2 2H2O + 213 kJ
5.2 L ? kJ
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 + 213 kJ
5.2 L ? kJ
5.2 L x 1 mole 22.4 L
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2 2H2O + 213 kJ
5.2 L ? kJ
5.2 L x 1 mole x 213 kJ 22.4 L 2 moles H2
5. How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP?
2H2 + O2 2H2O + 213 kJ
5.2 L ? kJ
5.2 L x 1 mole x 213 kJ = 25 kJ 22.4 L 2 moles H2
Home work
Worksheet # 3 page 131