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