1 / 44

Lesson 3 Percentage Yield and Energy

Lesson 3 Percentage Yield and Energy. 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

cbertrand
Télécharger la présentation

Lesson 3 Percentage Yield and Energy

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  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.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.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.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.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

  8. 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

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

  10. 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

  11. 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 2 mole AgNO3

  12. 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 311.9 g = 139.5 g 169.9 g 2 mole AgNO3 1 mole

  13. 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 %

  14. 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

  15. Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energyconsumed 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

  16. Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energyconsumed 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

  17. Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energyconsumed 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 2.02 g

  18. Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energyconsumed 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 2.02 g 2 mole H2

  19. Energy Calculations The energy term in a balanced equation can be used to calculate the amount of energyconsumed 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 kJ 2.02 g 2 mole H2

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

  30. 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

  31. 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?

  32. 2NaOH + 2Al +6H2O → 2NaAl(OH)4 + 3H2 3.00 M 50.0L ? Vol. What volume of sodium hydroxide solution is required? Note: use Molarity as your mole to volume conversion

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

  34. Remember:

  35. How to determine an unknown Molarity: • Ex 1: • 1. #moles KOH = M x vol • = 0.500mol x 0.025L • L • = 0.0125 mol

  36. How to experimentaly determine an Unknown Molarity: • Titration https://www.youtube.com/watch?v=sFpFCPTDv2w http://ocw.mit.edu/resources/res-5-0001-digital-lab-techniques-manual-spring-2007/videos/titration/

  37. 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.

  38. What mass of Br2 is produced when 25.0 grams of K2CrO7, 55.0 grams of KBr and 66.0 grams of H2SO4 are reacted according to the equation below? How many grams of each excess reactant will remain unreacted? • K2CrO7 + 6KBr +7H2SO44K2SO4 + Cr2(SO4)3 +3Br2 + 7H2O

  39. Masses based on KBr

  40. Masses in Excess

More Related