1 / 35

Mole Review

Mole Review. 1.) Calculate the number of moles in 60.4L of O 2 . 2.) How many moles are there in 63.2g of Cl 2 ?. 1 mol O 2. 60.4L O 2. = 2.7 mol O 2. 22.4L O 2. 1mol Cl 2. 63.2g Cl 2. = 0.903mol Cl 2. 70g Cl 2. Ch. 9. Math In Chemistry Stoichiometry. Proportional Relationships.

jerold
Télécharger la présentation

Mole Review

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. Mole Review 1.) Calculate the number of moles in 60.4L of O2. 2.) How many moles are there in 63.2g of Cl2? 1 mol O2 60.4L O2 = 2.7 mol O2 22.4L O2 1mol Cl2 63.2g Cl2 = 0.903mol Cl2 70g Cl2

  2. Ch. 9 Math In Chemistry Stoichiometry

  3. Proportional Relationships Tiny Tyke Tricycle Company F + S + 3W + H + 2P → FSW3HP2 Scheduled to make 640 tricycles. How many wheels should they order?

  4. Proportional Relationships 2 1/4 c. flour 1 tsp. baking soda 1 tsp. salt 1 c. butter 3/4 c. sugar 3/4 c. brown sugar 1 tsp vanilla extract 2 eggs 2 c. chocolate chips Makes 5 dozen cookies. I have 5 eggs. How many cookies can I make? Ratio of eggs to cookies 5 eggs 5 doz. 2 eggs = 12.5 dozen cookies

  5. Proportional Relationships Stoichiometry mass relationships between substances in a chemical reaction for example: you can determine the amount of a compound required to make another compound based on the mole ratio Mole Ratio indicated by coefficients in a balanced equation 2 Mg + O2 2 MgO 2 Moles of magnesium react with 1 mole of oxygen to form 2 moles of magnesium oxide.

  6. 2 Mg + O2 2 MgO What would be the mole ratio of magnesium to magnesium oxide? 2 : 2 Conversion factor = 2 mol Mg 2 mol MgO What would be the mole ratio of oxygen to magnesium? 1 : 2 Conversion factor = 1 mol O2 (Mole Ratio) 2 mol Mg

  7. Practice 5 F2 + 2NH3 N2F4 + 6HF 1. What is the mole ratio of NH3 to F2? Write the mole ratio as a conversion factor. 2. What is the mole ratio of HF to N2F4? Write the mole ratio as a conversion factor. 2:5 2mol NH3 5mol F2 6:1 6mol HF 1mol N2F4

  8. Stoichiometry Steps 1. Write a balanced equation Identify known & unknown. 3. Convert known to mole (if necessary), line up conversion factors. 4. Use Mole Ratio. 5. Convert moles to unknown unit (if necessary). 6. Calculate and write units. Mole ratio- get from equation Units of unknown known Mol of unknown Mol of known Mol of unknown

  9. Mole - Mole Stoichiometry Formula: known mol mol unknown mol known • Write the known and unknown. • Use the balanced equation to find the mole ratio. • Calculate.

  10. __S + __O2→ __SO3 Mole-Mole Examples #1 • Write the equation. • Balance the equation. • How many moles of SO3 are produced when there are 4.5 moles of S? • Known = • Unknown =

  11. 2C3H7OH + 9O2→ 6CO2 + 8H2O Mole-Mole Examples #2 Isopropyl alcohol (C3H7OH) burns in the air to this equation: • Write the equation. • Calculate the moles of oxygen needed to react with 3.40 moles of isopropyl alcohol.

  12. 2C3H7OH + 9O2→ 6CO2 + 8H2O Mole-Mole Examples #3 • Find the moles of water when 6.20 mol O2 reacts with C3H7OH.

  13. mol unknown Molar mass unknown 1 mol Known Known g Molar mass Known mol known 1 mol unknown Mol ratio- get from equation Mass-Mass Stoichiometry • Mass of reactants equals the mass of products, Law of Conservation of Mass • ONLY mass and atoms are conserved in every chemical reaction 1. Write the known and unknown. 2. Find the molar mass of the known and unknown substances. 3. Use mole and molar mass conversion factors from Ch. 7 and mole ratios from the balanced equation to solve.

  14. Mass-Mass Stoichiometry #1 The reaction of fluorine with ammonia produces dinitrogen tetrafluoride and hydrogen fluoride. 5F2 + 2 NH3→ N2F4 + 6HF How many grams of NH3 are required to produce 7.38g HF?

  15. Mass-Mass Stoichiometry #2 5F2 + 2 NH3→ N2F4 + 6HF How many grams of N2F4 can be produced from 265g F2?

  16. Mass-Mass Stoichiometry #3 2C2H2 + 5O2→ 4CO2 + 2H2O How many grams of oxygen are required to burn 52.0g C2H2?

  17. Volume-Volume Stoichiometry • Formula to use: Known (L) 1 mol known mol unknown 22.4 L unknown 1 22.4 L known mol known 1 mol unknown • Write the known and unknown. • Use mole and volume conversion factors from Ch. 7 and the mole ratios from the balanced equation to solve.

  18. C3H8 + 5O2→ 3CO2 + 4H2O Volume-Volume Example #4 If 25 liters of oxygen are consumed in the above reaction, how many liters of carbon dioxide are produced?

  19. Formulas to Use Mole - Mole Known mol of unknown mol of known Mol Ratio – from equation Mass - Mass mol of unknown Known g 1 mol known molar mass unknown mol of known 1 mol unknown molar mass known OR OR Other 6.02 × 1023 particles mol of unknown Known L 1 mol known 22.4 L unknown OR mol of known 1 mol unknown particles 22.4 L known 6.02 × 1023 particles

  20. Stoichiometry Problems • How many moles of KClO3 must decompose in order to produce 9 moles of oxygen gas? 2KClO3 2KCl + 3O2 Known: 9 moles O2 Unknown: moles KClO3 9 mol O2 2 mol KClO3 3 mol O2 = 6 mol KClO3

  21. Stoichiometry Problems • How many grams of KClO3 are required to produce 9.00 L of O2 at STP? 2KClO3 2KCl + 3O2 Known: 9.00 L O2 Unknown: g KClO3 9.00 L O2 1 mol O2 22.4 L O2 2 mol KClO3 3 mol O2 122 g KClO3 1 mol KClO3 = 32.68 g KClO3

  22. Stoichiometry Problems • How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3 2Ag + Cu(NO3)2 Known: 12.0 g Cu Unknown: g Ag 12.0 g Cu 1 mol Cu 64 g Cu 2 mol Ag 1 mol Cu 108 g Ag 1 mol Ag = 40.5 g Ag

  23. Stoichiometry Problems • How many grams of silver will be formed from 12.0 g copper? Cu + 2AgNO3 2Ag + Cu(NO3)2 Known: 12.0 g Cu Unknown: g Ag 12.0 g Cu 1 mol Cu 64 g Cu 2 mol Ag 1 mol Cu 108 g Ag 1 mol Ag = 40.5 g Ag

  24. Limiting Reactants/Reagents • Available Ingredients • 4 slices of bread • 1 jar of peanut butter • 1/2 jar of jelly • Limiting Reactant/Reagents • bread • Excess Reactants/Reagents • peanut butter and jelly

  25. Limiting Reactants/Reagents • Limiting Reactant/Reagent • used up in a reaction • determines the amount of product • Excess Reactant/Reagent • added to ensure that the other reactant is completely used up • cheaper & easier to recycle

  26. To Determine Limiting Reagents 1. Write a balanced equation. 2. For each reactant, calculate the amount of product formed. 3. Smaller answer indicates: • limiting reactant • actual amount of product

  27. Limiting Reagents 79.1 g of zinc react with 0.90 L of HCl. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L ? L

  28. Limiting Reagents Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L ? L 79.1 g Zn 1 mol Zn 65 g Zn 1 mol H2 1 mol Zn 22.4 L H2 1 mol H2 = 27.26 L H2

  29. Limiting Reagents Zn + 2HCl  ZnCl2 + H2 79.1 g 0.90 L ? L 0.90 L HCl 1 mol HCl 22.4 L HCl 1 mol H2 2 mol HCL 22.4 L H2 1 mol H2 = 0.45 L H2

  30. Limiting Reagents Zn: 27.26 L H2HCl: 0.45 L H2 Limiting reagent: HCl Excess reagent: Zn

  31. Percent Yield Percent yield- the ratio of the actual yield to the theoretical yield Actual yield- the amount of product formed when a reaction is carried out in the laboratory Theoretical yield- the calculated amount of product formed during a reaction (mathematical calculation used to make answer keys)

  32. Percent Yield measured in lab calculated on paper

  33. Percent Yield • When 45.8 g of K2CO3 react with excess HCl, 46.3 g of KCl are formed. Calculate the theoretical and % yields of KCl. K2CO3 + 2HCl  2KCl + H2O + CO2 45.8 g ? g actual: 46.3 g

  34. B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield: 45.8 g ? g actual: 46.3 g 45.8 g K2CO3 1 mol K2CO3 138 g K2CO3 2 mol KCl 1 mol K2CO3 74 g KCl 1 mol KCl = 49.12 g KCl

  35. 46.3 g 49.12 g B. Percent Yield K2CO3 + 2HCl  2KCl + H2O + CO2 Theoretical Yield = 49.12 g KCl 45.8 g 49.12 g actual: 46.3 g  100 = 94.3% % Yield =

More Related