Lesson 3 Percentage 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.

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Lesson 3 Percentage 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 Fe 2 O 3, if the percentage yield is 75.0%? 2Fe 2 O 3 +3C  4Fe+3CO g? g 100. g Fe 2 O 3

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 Fe 2 O 3, if the percentage yield is 75.0%? 2Fe 2 O 3 +3C  4Fe+3CO g? g 100. g Fe 2 O 3 x 1 mole 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 Fe 2 O 3, if the percentage yield is 75.0%? 2Fe 2 O 3 +3C  4Fe+3CO g? g 100. g Fe 2 O 3 x 1 mole x 4 mole Fe g 2 mole Fe 2 O 3

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 Fe 2 O 3, if the percentage yield is 75.0%? 2Fe 2 O 3 +3C  4Fe+3CO g? g 100. g Fe 2 O 3 x 1 mole x 4 mole Fe x 55.8 g g 2 mole Fe 2 O 3 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 Fe 2 O 3, if the percentage yield is 75.0%? 2Fe 2 O 3 +3C  4Fe+3CO g? g 100. g Fe 2 O 3 x 1 mole x 4 mole Fe x 55.8 g x = 52.4 g g 2 mole Fe 2 O 3 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 AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield g actual yield 2AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 152 g ? g 152. g AgNO 3 x 1 mole g

2.In an experiment 152. g of AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield g actual yield 2AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 152 g ? g 152. g AgNO 3 x 1 mole x 1 Ag 2 SO g 2 mole AgNO 3

2.In an experiment 152. g of AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield g actual yield 2AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 152 g ? g 152. g AgNO 3 x 1 mole x 1 Ag 2 SO 4 x g = g g 2 mole AgNO 3 1 mole

2.In an experiment 152. g of AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield g actual yield 2AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 152 g ? g 152. g AgNO 3 x 1 mole x 1 Ag 2 SO 4 x g = g g 2 mole AgNO 3 1 mole % yield = 75.1 x 100 %

2.In an experiment 152. g of AgNO 3 is used to make 75.1 g of Ag 2 SO 4(s). Calculate the percentage yield g actual yield 2AgNO 3(aq) + Na 2 SO 4(aq)  Ag 2 SO 4(s) + 2NaNO 3(aq) 152 g ? g 152. g AgNO 3 x 1 mole x 1 Ag 2 SO 4 x g = g g 2 mole AgNO 3 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 H 2 ? 213 kJ +2H 2 O  2H 2 +O 2 ? kJ25.4 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 H 2 ? 213 kJ +2H 2 O  2H 2 +O 2 ? kJ25.4 g 25.4 g H 2

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 H 2 ? 213 kJ +2H 2 O  2H 2 +O 2 ? kJ25.4 g 25.4 g H 2 x 1 mole 2.02 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 H 2 ? 213 kJ +2H 2 O  2H 2 +O 2 ? kJ25.4 g 25.4 g H 2 x 1 mole x 213 kJ 2.02 g2 mole H 2

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 H 2 ? 213 kJ +2H 2 O  2H 2 +O 2 ? kJ25.4 g 25.4 g H 2 x 1 mole x 213 kJ = 1.34 x 10 3 kJ 2.02 g2 mole H 2

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? 2H 2 +O 2  2H 2 O +213 kJ ? Molecules452 kJ 452 kJ

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? 2H 2 +O 2  2H 2 O +213 kJ ? Molecules452 kJ 452 kJ x 2 moles H kJ

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? 2H 2 +O 2  2H 2 O +213 kJ ? Molecules452 kJ 452 kJ x 2 moles H 2 x 6.02 x molecules 213 kJ 1 mole

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? 2H 2 +O 2  2H 2 O +213 kJ ? Molecules452 kJ 452 kJ x 2 moles H 2 x 6.02 x molecules = 2.55 x molecs 213 kJ 1 mole

4.How many molecules of H 2 can be produced when 452 kJ of energy if consumed? 2H 2 +O 2  2H 2 O +213 kJ ? Molecules452 kJ 452 kJ x 2 moles H 2 x 6.02 x molecules = 2.55 x 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? 2H 2 +O 2  2H 2 O +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? 2H 2 +O 2  2H 2 O +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? 2H 2 +O 2  2H 2 O +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? 2H 2 +O 2  2H 2 O +213 kJ 5.2 L? kJ 5.2 L x 1 mole x 213 kJ 22.4 L 2 moles H 2

5.How much energy is produced by an explosion of a 5.2 L balloon full of hydrogen at STP? 2H 2 +O 2  2H 2 O +213 kJ 5.2 L? kJ 5.2 L x 1 mole x 213 kJ = 25 kJ 22.4 L 2 moles H 2 Home work Worksheet # 3page 131