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the Mole !

It’s time to learn about. the Mole !. Stoichiometry: Limiting Reagents At the conclusion of our time together, you should be able to:. Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction

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the Mole !

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  1. It’s time to learn about . . . the Mole !

  2. Stoichiometry: Limiting ReagentsAt the conclusion of our time together, you should be able to: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction

  3. OK, you guys, you really need to follow my lead on this!!!

  4. Limiting Reactant: Cookies 1 cup butter 1/2 cup white sugar 1 cup packed brown sugar 1 teaspoon vanilla extract 2 eggs 2 1/2 cups all-purpose flour 1 teaspoon baking soda 1 teaspoon salt 2 cups semisweet chocolate chips Makes 3 dozen If we had the specified amount of all ingredients listed and 4 extra eggs, could we make 4 dozen cookies? If not, what limits us?? What if we only had one egg, could we make 3 dozen cookies???

  5. Limiting Reactant • Most of the time in chemistry we have more of one reactant than we need to completely use up other reactant. • That reactant is said to be in excess (there is too much). • The other reactant limits how much product we get. Once it runs out, the reaction ’s. This is called the limiting reactant.

  6. LimitingReactant Limiting Reactant: Example • 10.0 g of aluminum reacts with 35.0 grams of chlorine gas to produce aluminum chloride. Which reactant is limiting, which is in excess, and how much product is produced? 2 Al + 3 Cl2 2 AlCl3 • Start with Al: • Now Cl2: 10.0 g Al 1 mol Al 2 mol AlCl3 133.33 g AlCl3 26.98 g Al 2 mol Al 1 mol AlCl3 = 49.4 g AlCl3 35.0 g Cl2 1 mol Cl2 2 mol AlCl3 133.33 g AlCl3 70.90 g Cl2 3 mol Cl2 1 mol AlCl3 = 43.9 g AlCl3

  7. LR Example Continued • We get 49.4 g of aluminum chloride from the given amount of aluminum, but only 43.9 g of aluminum chloride from the given amount of chlorine. Therefore, chlorine is the limiting reactant. Once the 35.0 g of chlorine is used up, the reaction comes to a complete .

  8. Why Don't Blind People Like To Sky Dive? Because It Scares The Dog.

  9. Limiting Reactant Practice #5 • 3.00 g of magnesium reacts with 2.20 g of oxygen. Calculate which reactant is in excess and how much product is made. • A little quicker way to do this is to pick one reactant and determine what amount of the other reactant is needed to completely react the first.

  10. x 32.00 g O2 1 mol O2 x 1mol O2 x 1 mol Mg 2 mol Mg 24.31 g Mg Calculate the mass in grams of oxygen required to react completely with 3.00 g of magnesium. Mg+ O2 MgO 2 Mg+ O2 2 MgO 3.00 g Mg = 1.97 g O2

  11. Finding the Amount of Excess • To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have 2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess. • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

  12. Finding the Amount of Excess • Can we find the amount of excess oxygen in the previous problem? • To completely react 3.00 g of magnesium, we would need 1.97 g of oxygen. Since we have 2.20 g of oxygen, the magnesium limits the amount of product we can make. Oxygen is in excess. • 2.20 g O2 available – 1.97 g O2 used = • 0.23 g O2 excess

  13. x 40.31 g MgO 1 mol MgO x 2 mol MgO x 1 mol Mg 2 mol Mg 24.31 g Mg Calculate the mass in grams of magnesium oxide produced when 3.00 g of magnesium (the limiting reagent) is burned in excess oxygen. Mg+ O2 MgO 2 Mg+ O2 2 MgO 3.00 g Mg = 4.97 g MgO

  14. Gee, I wonder what’s going to happen next??

  15. Stoichiometry: Limiting ReagentsAt the conclusion of our time together, you should be able to: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction

  16. Limiting Reactant: Recap • You can recognize a limiting reactant problem because there is MORE THAN ONE GIVEN AMOUNT. • Pick a reactant (A), convert grams to moles, and compare it to the grams needed of the other reactant (B). • If the answer for B is higher than what you have in the problem, • You don’t have enough B, therefore B is limiting. • If the answer for B is lower than what you have in the problem, • You have enough B, therefore A is limiting. • If you have enough B, subtract your answer for B from what you have in the problem = excess B.

  17. Stoichiometry: Limiting ReagentsLet’s see if you can: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction

  18. Interesting Picture:

  19. What mass of ZnO is formed when 20.0 g of MoO3 is reacted with 10.0 g of Zn? x 3 mol Zn x 1 mol MoO3 x 65.39 g Zn 2 mol MoO3 143.94 g MoO3 1 mol Zn 3 Zn + 2 MoO3 Mo2O3 + 3 ZnO 20.0 g MoO3 = 13.6 g Zn

  20. What mass of ZnO is formed when 20.0 g of MoO3 is reacted with 10.0 g of Zn?We need 13.6 g of Zn so Zn is limiting! x 3 mol ZnO x 1 mol Zn x 81.39 g ZnO 3 mol Zn 65.39 g Zn 1 mol ZnO 3 Zn + 2 MoO3 Mo2O3 + 3 ZnO 10.0 g Zn = 12.4 g ZnO

  21. Stoichiometry: Limiting ReagentsLet’s see if you can: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction

  22. x 163.94 g Na3PO4 1 mol Na3PO4 x 1 mol NH4NO3 x 1 mol Na3PO4 80.06 g NH4NO3 3 mol NH4NO3 Another Limiting Reagent Worksheet #1 3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3 a. Which is limiting? 30.0 g NH4NO3 = 20.5 g Na3PO4

  23. x 149.12 g (NH4)3PO4 1 mol (NH4)3PO4 x 1 mol NH4NO3 x 1 mol (NH4)3PO4 80.06 g NH4NO3 3 mol NH4NO3 30.0 g NH4NO3 = 18.6 g (NH4)3PO4 a. Which is limiting? NH4NO3 3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3 b. Maximum amount of each product?

  24. x 85.00 g NaNO3 1 mol NaNO3 x 1 mol NH4NO3 x 3 mol NaNO3 80.06 g NH4NO3 3 mol NH4NO3 30.0 g NH4NO3 = 31.9 g NaNO3 a. Which is limiting? NH4NO3 3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3 b. Maximum amount of each product?

  25. - 20.5 g NH3PO4 50.0 g Na3PO4 = 29.5 g NH3PO4 c. How much of the other reagent is left over?? 3 NH4NO3 + Na3PO4  (NH4)3PO4 + 3 NaNO3

  26. Stoichiometry: Limiting ReagentsLet’s see if you can: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction

  27. FeS Lab Problem: In the lab where you combined iron (II) with sulfur, if 5.00 g of iron is combined with 5.00 g of sulfur, what is the limiting reagent and how much excess reagent is left? x 1 mol Fe x 1 mol S x 32.07 g S 55.85 g Fe I mol Fe 1 mol S Fe + S  FeS 5.00 g Fe = 2.87 g S

  28. When 5.00 g of Fe is used, I would need 2.87 grams of S to completely react the Fe. Since I have 5.00 g of S, I have plenty of S and therefore, the Fe is limiting. 2.87 g S used 5.00 g S – 2.87 g S used = 2.13 g excess S

  29. Stoichiometry: Limiting Reagents with Percent YieldAt the conclusion of our time together, you should be able to: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction Determine the percent yield for a problem

  30. What is the Percent Yield of a Chemical Reaction? Actual Product Produced Theoretical Product that Should be Produced X 100 = Percent Yield

  31. Limiting Reactant Practice • 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate which reactant is limiting, how much excess reactant is available and how much product is made. • Determine the percent yield if 10.5 g of aluminum iodide is actually produced.

  32. x 253.80 g I2 1 mol I2 x 3 mol I2 x 1 mol Al 2 mol Al 26.98 g Al Calculate the mass in grams of iodine required to react completely with 15.0 g of aluminum. Al + I2 AlI3 2Al + 3 I2 2 AlI3 15.0 g Al = 212 g I2

  33. Finding the Amount of Excess • To completely react 15.0 g of aluminum, we would need 212 g of iodine. Since we only have 15.0 g of iodine, the iodine limits the amount of product we can make. • By calculating the amount of the excess reactant needed to completely react with the limiting reactant, we can subtract that amount from the given amount to find the amount of excess.

  34. Finding Excess Practice • 15.0 g of aluminum reacts with 15.0 g of iodine. Calculate the excess of aluminum. 2Al + 3 I2 2 AlI3 Always start with the limiting reactant: 15.0 g I2 1 mol I2 2 mol Al 26.98 g Al 253.80 g I2 3 mol I2 1 mol Al = 1.06 g Al USED! 15.0 g Al – 1.06 g Al = 13.9 g Al EXCESS Note that we started with the limiting reactant! Once you determine the LR, you should only use it! Given amount of excess reactant Amount of excess reactant actually used

  35. x 407.68 g AlI3 1 mol I2 x 1 mol I2 x 2 mol AlI3 253.80 g I2 3 mol I2 Calculate the mass in grams of aluminum iodide that would be produced in this reaction. Al + I2 AlI3 2Al + 3 I2 2 AlI3 15.0 g I2 = 16.1 g AlI3

  36. 10.5 g AlI3 x 100 16.1 g AlI3 Determine the percent yield if 10.5 g of aluminum iodide is produced. 16.1 g of AlI3 should have been produced. What is the percent yield? 2Al + 3 I2 2 AlI3 = 65.2 % Yield

  37. What's The Difference Between a Bad Golfer And a Bad Skydiver? A Bad Golfer Goes, Whack, Dang!A Bad Skydiver Goes Dang! Whack.

  38. Stoichiometry: Limiting Reagents with Percent YieldLet’s see if you can: Determine the limiting reagent Use the limiting reagent to determine the amount of product produced in a reaction Determine the excess amount(s) in the reaction Determine the percent yield for a problem

  39. Percent Yield #1 • How many grams of antimony(III) iodide would be produced? • Determine the percent yield if 118.00 g of antimony(III) iodide is actually produced.

  40. x 502.46 g SbI3 1 mol SbI3 x 1 mol I2 x 2 mol SbI3 253.80 g I2 3 mol I2 How many grams of antimony(III) iodide would be produced? Sb + I2 SbI3 2Sb + 3 I2 2 SbI3 98.60 g I2 = 130.1 g SbI3

  41. x 118.00 g SbI3 x 100 130.1 g SbI3 Determine the percent yield if 118.00 g of antimony (III) iodide is produced. 130.1 g of Sbl3 should have been produced. What is the percent yield? 2Sb + 3 I2 2 SbI3 = 90.70 % Yield

  42. Way to Kick Butt!!!!

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