1. Chapter 12 Stoichiometry Anything in black letters = write it in your notes (‘knowts’) 12.1 – The Arithmetic of Equations 12.2 – Chemical Calculations 12.3 – Limiting Reagent & Percent Yield

2. 12.1 – The Arithmetic of Equations Quantitative study of chemical rxns. In this chapter we will be asking ‘how much?’ Stoichiometry -

3. The coefficients of a balanced chemical equation represent the number of moles that are reacting or produced. 2H 2 + O 2  2H 2 O

4. N 2 + 3H 2  2NH 3 1  ( ) + 3  ( )  2  ( ) 6.02  10 23 molecules N 2 6.02  10 23 molecules H 2 6.02  10 23 molecules NH 3 1 N 2 molecule + 3 H 2 molecules  2 NH 3 molecules It is even more practical to talk about moles… It is not practical to talk about single molecules; instead use a larger number of molecules… 1 mol of N 2 + 3 mol of H 2  2 mol of NH 3

5. The bike example… For simplicity, say a bike requires only two things: 1 frame and 2 wheels. 2wheels + 1frame  1bike What are the coefficients here? What do they tell you?

6. How many frames would be needed to ‘react’ completely with 20 wheels? How many bikes could be produced from 4 wheels and 560 frames? What is the limiting reactant here?  + 2wheels + 1frame  1bike

7. How many bikes would be produced from 23.7 kg of wheels and 80.1 kg of frames? What is needed in order to solve the above question? Always convert to a number of things (mol) first!  + 2wheels + 1frame  1bike

8. ASSIGNMENT: Chapter 12 Worksheet #1

9. 12.2 – Chemical Calculations A mole ratio is a conversion factor that comes from the coefficients of a balanced chemical equation.

10. N 2 (g) + 3H 2 (g)  2NH 3 (g) 2 mol NH 3 1 mol N 2 3 mol H 2 2 mol NH 3 Write the three mole ratios that can be written from this balanced equation… 2 mol NH 3 1 mol N 2 3 mol H 2 2 mol NH 3 These are equivalent ratios, just upside down…

11. N 2 (g) + 3H 2 (g)  2NH 3 (g) Mole ratios are used to convert from a given number of moles of a reactant or product to moles of a different reactant or product. Example 1 : How many moles of NH 3 are produced when 0.60 mol of N 2 reacts with excess H 2 ? 0.60 mol N 2  2 mol NH 3 1 mol N 2 = 1.2 mol NH 3

12. N 2 (g) + 3H 2 (g)  2NH 3 (g) Example 2: Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen.  2 mol NH 3 3 mol H 2 5.40 g H 2  1 mol H 2 2.0 g H 2 17.0 g NH 3 1 mol NH 3  Given quantity Change given unit to moles Mole ratio Change moles to grams = 31 g NH 3

13. Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O 2 (g)  P 4 O 10 (s) What mass of phosphorus will be needed to produce 3.25 mol of P 4 O 10 ? 3.25 mol P 4 O 10  31.0 g P 1 mol P = 403 g P 1 mol P 4 O 10 4 mol P 

14. How do you get better at something? Did you tie your shoe correctly the very first time? Did you recite the alphabet correctly the first time? Try the greek alphabet

15. How to Solve Stoichiometric Problems - Streamlined 1.Convert given # into moles, if it isn’t already 2.Multiply by the mole ratio conversion factor 3.Convert from moles of substance into desired unit if necessary.

16. What mass of frames would be needed to ‘react’ completely with 3060 g wheels  + 2wheels + 1frame  1bike 45 g/wheel 27 g/frame 117 g/bike 3060 g wheels  1 frame 2 wheel = 918 g frames 45 g wheels 1 wheel  27 g frame 1 frame 

17. ASSIGNMENT: Chapter 12 #11-16; 21-25 (p.391-398)

18. ASSIGNMENT: Chapter 12 #45-50(p.411)

25. 12.3 – Limiting Reagent & Percent Yield The substance that is completely used up in a chemical rxn is called the limiting reagent. A reagent is also known as a reactant The substance that is NOT completely used up (and partially remains) is the excess reactant.

26. Example: Copper reacts with sulfur to form copper(I) sulfide. What is the limiting reagent when 80.0 grams of Cu react with 25.0 g S? 2Cu + S  Cu 2 S 80.0 g Cu  63.5 g Cu 1 mol Cu 1. Calculate the amount of one reactant required to react with the other. 2. Compare the given amount to the required amount.  2 mol Cu 1 mol S  32.06 g S = 21.2 g S required amount.

27. 0.630 mol S is needed to react with 1.26 mol Cu, there was 0.779 mol S given in the question. Sulfur is in excess, so Cu is the limiting reagent 3. Compare the given amount to the required amount.

28. It doesn’t matter which reactant you use. If you used the actual amount of moles of S to find the amount of copper needed, then you would still identify copper as the limiting reagent.

29. You Try It! 2Fe + O 2 + 2H 2 O  2Fe(OH) 2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 2. Compare the given amount to the required amount. 1. Calculate the amount of one reactant required to react with the other.

30. The theoretical yield is the calculated amount of product that could be formed from given amounts of reactants; it is a the maximum amount. The amount of product that actually forms when the reaction is carried out in the laboratory is called the actual yield; it is usually lower than the theoretical yield percent yield = actual yield theoretical yield  100%

31. CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 You Try It! What is the percent yield of CCl 4 if 617 kg is produced from the reaction of 312 kg of CS 2 ? 3.12  10 5 g CS 2    76.142 g CS 2 1 mol CS 2 1 mol CCl 4 1 mol CS 2 153.81 g CCl 4 1 mol CCl 4 = 6.30  10 5 g CCl 4 = 630 kg CCl 4 Percent yield =  100% = 97.9% 617 kg CCl 4 630 kg CCl 4

32. ASSIGNMENT: Chapter 12 #26-38 (p.402 - 408)

33. 1 ream of paper = 500 sheets 1 ream of copy paper = 5 lbs 1 ream of card-stock paper = 12 lbs (1) 10 lbs of copy paper = ________ reams (2) 1.31 lbs of copy paper = _______ reams a paper analogy…

34. 1 ream of paper = 500 sheets 1 ream of copy paper = 5 lbs 1 ream of card-stock paper = 12 lbs (3) 0.23 reams card stock paper = _____ lbs (4) 5.4 lbs of card stock paper = _______ sheets