Chapter 12 Stoichiometry

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

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

3. The coefficients of a balanced chemical equation represent the number of moles that are reacting or produced. 2H2 + O2 2H2O

4. 1 ( ) + 3 ( ) 2 ( ) 6.02  1023 molecules N2 6.02  1023 molecules H2 6.02  1023 molecules NH3 N2 + 3H2 2NH3 1N2 molecule + 3 H2 molecules  2NH3 molecules It is not practical to talk about single molecules; instead use a larger number of molecules… It is even more practical to talk about moles… 1 mol of N2 + 3 mol of H2 2 mol of NH3

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. +  2wheels + 1frame  1bike 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?

7. +  2wheels + 1frame  1bike 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!

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. 1 mol N2 3 mol H2 2 mol NH3 1 mol N2 3 mol H2 2 mol NH3 3 mol H2 1 mol N2 2 mol NH3 1 mol N2 2 mol NH3 3 mol H2 N2(g) + 3H2(g)  2NH3(g) Write the three mole ratios that can be written from this balanced equation… These are equivalent ratios, just upside down…

11. 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. N2(g) + 3H2(g)  2NH3(g) Example 1 : How many moles of NH3 are produced when 0.60 mol of N2 reacts with excess H2? 2 mol NH3  = 1.2 mol NH3 0.60 mol N2 1 mol N2

12. 1 mol H2 2 mol NH3 17.0 g NH3 5.40 g H2   2.0 g H2 3 mol H2 1 mol NH3 Change moles to grams Given quantity Change given unit to moles Mole ratio Example 2: N2(g) + 3H2(g)  2NH3(g) Calculate the number of grams of NH3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. = 31 g NH3

13. Phosphorus burns in air to produce a phosphorus oxide in the following reaction: 4P(s) + 5O2(g)  P4O10(s) What mass of phosphorus will be needed to produce 3.25 mol of P4O10? 4 mol P 31.0 g P   3.25 mol P4O10 = 403 g P 1 mol P4O10 1 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 • Convert given # into moles, if it isn’t already • Multiply by the mole ratio conversion factor • Convert from moles of substance into desired unit if necessary.

16.  + 2wheels + 1frame  1bike 45 g/wheel27 g/frame117 g/bike What mass of frames would be needed to ‘react’ completely with 3060 g wheels 1 wheel 1 frame 27 g frame    3060 g wheels 2 wheel 1 frame 45 g wheels = 918 g frames

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  Cu2S 1. Calculate the amount of one reactant required to react with the other. 1 mol Cu 1 mol S 32.06 g S    80.0 g Cu 63.5 g Cu 2 mol Cu 1 mol S = 21.2 g S required amount. 2. Compare the given amount to the required amount.

27. 3. Compare the given amount to the required amount. 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

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 + O2 + 2H2O  2Fe(OH)2 If 7.0 g of iron and 9.0 g of water are available to react, which is the limiting reagent? 1. Calculate the amount of one reactant required to react with the other. 2. Compare the given amount to the required amount.

30. actual yield percent yield =  100% theoretical yield 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

31. 1 mol CS2 1 mol CCl4 153.81 g CCl4 3.12  105 g CS2    76.142 g CS2 1 mol CS2 1 mol CCl4 617 kg CCl4 630 kg CCl4 Percent yield =  100% = 97.9% You Try It! What is the percent yield of CCl4 if 617 kg is produced from the reaction of 312 kg of CS2? CS2 + 3Cl2 CCl4 + S2Cl2 = 6.30  105 g CCl4 = 630 kg CCl4

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

33. a paper analogy… 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

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