  # Chapter 12 Stoichiometry 1 Section 12.1 The Arithmetic of - PowerPoint PPT Presentation

## Chapter 12 Stoichiometry 1 Section 12.1 The Arithmetic of Equations 2 Cookies and ChemistryHuh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction

1. Chapter 12 Stoichiometry 1

2. Section 12.1 The Arithmetic of Equations 2

3. Cookies and Chemistry…Huh!?!? Just like chocolate chip cookies have recipes, chemists have recipes as well Instead of calling them recipes, we call them reaction equations Furthermore, instead of using cups and teaspoons, we use moles Lastly, instead of eggs, butter, sugar, etc. we use chemical compounds as ingredients 3

4. Chemistry Recipes Looking at a reaction tells us how much of something you need to react with something else to get a product (like the cookie recipe) Be sure you have a balanced reaction before you start! Example: 2 Na + Cl 2 → 2 NaCl This reaction tells us that by mixing 2 moles of sodium with 1 mole of chlorine we will get 2 moles of sodium chloride What if we wanted 4 moles of NaCl? 10 moles? 50 moles? 4

5. 2H 2 + O 2 → 2H 2 O ◼ Two molecules of hydrogen and one molecule of oxygen form two molecules of water. ◼ 2 A l 2 O 3 →  A l + 3O 2 2 formula units Al 2 O 3 form 4 atoms Al and 3 molecules O 2 Now try this: 2Na + 2H 2 O → 2NaOH + H 2 5

6. ◼ Write the balanced reaction for hydrogen gas reacting with oxygen gas. 2 H 2 + O 2 → 2 H 2 O ➢ How many moles of reactants are needed? ➢ What if we wanted 4 moles of water? ➢ What if we had 3 moles of oxygen, how much hydrogen would we need to react and how much water would we get? ➢ What if we had 50 moles of hydrogen, how much oxygen would we need and how much water produced? 6

7. Section 12.2 Chemical Calculations 7

8. Mole Ratios These mole ratios can be used to calculate the moles of one chemical from the given amount of a different chemical Example: How many moles of chlorine are needed to react with 5 moles of sodium (without any sodium left over)? Step 1: Write the chemical equation. Step 2: Balance the equation. 2 Na + Cl 2 → 2 NaCl Step 3: (Write the mole ratio) 2 mol 1 mol (Write the given) Step 4: 5 mol x  5 1 Step 5: = = x 2 . 5 mol Cl 2 2 8

9. Mole - Mole Conversions Example: ◼ How many moles of sodium chloride will be produced if you react 2.6 moles of chlorine gas with an excess (more than you need) of sodium metal? 2 Na + Cl 2 2 NaCl 1 mol 2 mol x 2.6 mol  2 . 6 2 = = x 5 . 2 mol NaCl 1 9

10. Mole - Mole Conversions Example: ◼ How many moles of O 2 are produced when 3.34 moles of Al 2 O 3 decompose? 2 Al 2 O 3 → Al +  O 2 3 2 mol 3 mol 3.34 mol x  3 . 34 3 = = x 5 . 01 mol O 2 2 10

11. Mole - Mass Conversions Most of the time in chemistry, the amounts are given in grams instead of moles We still go through moles and use the mole ratio , but now we also use molar mass to get to grams 11

12. ◼ Example: How many grams of chlorine are required to react completely with 5.00 moles of sodium to produce sodium chloride? 2 Na + Cl 2 → 2 NaCl 2 mol 1 mol 5 mol x  5 1 = = x 2 . 5 mol Cl 2 2 Change moles to grams: No. of moles X molar mass 2.5 X 71 = 178 g Cl 2 12

13. Example: ◼ Calculate the mass in grams of Iodine required to react completely with 0.50 moles of aluminum. 3 2 I 2 + Al → AlI 3 2 3 mol 2 mol 0.5 mol x  3 0 . 5 = = x 0 . 75 mol I 2 2 Change moles to grams: No. of moles X molar mass 0.75 X 254 = 191 g I 2 13

14. Mass – Mole Conversions We can also start with mass and convert to moles of product or another reactant We use molar mass and the mole ratio to get to moles of the compound of interest 14

15. Example: ◼ Calculate the number of moles of ethane (C 2 H 6 ) needed to produce 10.0 g of water 2 C 2 H 6 + 7 O 2 → 4 CO 2 + 6 H 2 O 2 mol 6 mol 10 g x But before making cross multiplication, change 10 grams to moles 10 / 18.02 = 0.555 mol  0 . 555 2 = = x 0 . 185 mol C 2 H 6 6 15

16. Mass – Mass Conversion Example: Calculate how many grams of ammonia are produced ◼ when you react 2.00g of nitrogen with excess hydrogen. N 2 + H 2 → 2 NH 3 3 1 mol 2 mol 2 g x But before making cross multiplication, change 2 grams to moles 2 / 28 = 0.0714 mol  0 . 0714 2 = = x 0 . 143 mol NH 3 1 Change moles to grams: No. of moles X molar mass 0.143 X 17.03 = 2.44 g NH 3 16

17. Example: ◼ Calculate how many grams of oxygen are required to make 10.0 g of aluminum oxide  O 2 → Al + 3 Al 2 O 3 2 3 mol 2 mol 10 g x But before making cross multiplication, change 10 grams to moles 10 / 102 = 0.098 mol  0 . 098 3 = = x 0 . 147 mol O 2 2 Change moles to grams: No. of moles X molar mass 0.147 X 32 = 4.7 g O 2 17

18. Example: • If 3.84 moles of C 2 H 2 are burned, how many moles of O 2 are needed? (9.6 mol) 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 2 H 2 O 2 mol 5 mol 3.84 mol x  3 . 84 5 = = x 9 . 6 mol O 2 2 18

19. Example: • How many moles of C 2 H 2 are needed to produce 8.95 mole of H 2 O? (8.95 mol) 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 2 H 2 O 2 mol 2 mol 8.95 mol x  8 . 95 2 = = x 8 . 95 mol C 2 H 2 2 19

20. Example: • If 2.47 moles of C 2 H 2 are burned, how many moles of CO 2 are formed? (4.94 mol) 2 C 2 H 2 + 5 O 2 → 4 CO 2 + 2 H 2 O 2 mol 4 mol 2.47 mol x  2 . 47 4 = = x 4 . 94 mol CO 2 2 20

21. Mass – Mass Problem: 6.50 grams of aluminum reacts with an excess of oxygen. How many grams of aluminum oxide are formed? 4 Al + 3 O 2 → 2Al 2 O 3 4 mol 2 mol 6.5 g x But before making cross multiplication, change 6.5 grams to moles 6.5 / 27 = 0.241 mol Al  0 . 241 2 = = x 0 . 121 mol Al 2 O 3 4 Change moles to grams: No. of moles X molar mass 0.121 X 102 = 12.3 g Al 2 O 3 21

22. Another example: If 10.1 g of Fe are added to a solution of Copper (II) Sulfate, how many grams of solid copper would form? 2 Fe + 3 CuSO 4 → Fe 2 (SO 4 ) 3 + 3 Cu 2 mol 3 mol 10.1 g x before making cross multiplication, change 10.1 grams to moles 10.1 / 55.8 = 0.181 mol Fe  0 . 181 3 = = x 0 . 272 mol Cu 2 Change moles to grams: No. of moles X molar mass 0.272 X 63.5 = 17.3 g Cu 22

23. Volume – Volume Calculations: ◼ How many liters of CH 4 at STP are required to completely react with 17.5 L of O 2 ? CH 4 + 2 O 2 → CO 2 + 2 H 2 O 1 mol 2 mol 17.5 L x before making cross multiplication, change 17.5 L to moles 17.5 / 22.4 = 0.781 mol O 2  0 . 781 1 = = x 0 . 391 mol CH 4 2 Change moles to liters: No. of moles X 22.4 0.391 X 22.4 = 8.76 L CH 4 23

24. Section 12.3 Limiting Reactant and Yield 24

25. 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, could we make 4 dozen cookies? What if we had 6 eggs and twice as much of everything else, could we make 9 dozen cookies? What if we only had one egg, could we make 3 dozen cookies?

26. 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.

27. Limiting Reactant ◼ To find the correct answer, we have to try all of the reactants. We have to calculate how much of a product we can get from each of the reactants to determine which reactant is the limiting one. ◼ The lower amount of a product is the correct answer. ◼ The reactant that makes the least amount of product is the limiting reactant. Once you determine the limiting reactant, you should ALWAYS start with it! ◼ Be sure to pick a product! You can’t compare to see which is greater and which is lower unless the product is the same!

28. Example ◼ 10.0g 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 A l + 3 Cl 2 2 AlCl 3 ◼ Start with Al: 2 mol 2 mol 10 g x But before making cross multiplication, change 10 grams to moles 10 / 27 = 0.370 mol  0 . 370 2 = = 0.370 x 133.5 = 49.4 g x 0 . 370 mol A l C l 3 2 ◼ Now Cl 2 : 3 mol 2 mol 35 g x But before making cross multiplication, change 35 grams to moles 35 / 71 = 0.493 mol  0 . 493 2 = = 0.329 x 133.5 = 43.9 g x 0 . 329 mol A l C l 3 3

29. 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 33.6 g of chlorine is used up, the reaction comes to a complete .

30. Practice ◼ If 10.6 g of copper reacts with 3.83 g sulfur, how many grams of product (cuprous sulfide) will be formed? 30

31. ◼ 15.0 g of potassium reacts with 15.0 g of iodine. Calculate which reactant is limiting and how much product is made. 31

32. Finding the Amount of 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. ◼ Can we find the amount of excess potassium in the previous problem?

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