1. 4/15/16 You need a calculator. Find your periodic table. Find 9.3 notes: Limiting Reactants & Percent Yield Grab TWO handouts on counter. Listen for homework assignment at the end of class. Check the calendar for EV-ER-Y-TH-IN-G!

2. Bellwork Use this equation to answer the following questions 4Al + 3O 2  2Al 2 O 3 1.Which reactant is limiting if 0.32 g Al and 0.26 g O 2 are available? 2.How many grams of product are formed? 3.What mass of excess reactant is remaining?

3. Bellwork 4/13 You need a calculator and a periodic table. 1.What is the limiting reactant in a reaction? 2.Zinc citrate, Zn 3 (C 6 H 5 O 7 ) 2, is an ingredient in toothpaste. It is synthesized by the reaction: 3ZnCO3 + 2C 6 H 8 O 7  Zn 3 (C 6 H 5 O 7 ) 2 + 3H 2 O + CO 2 If there is 1 mol of ZnCO 3 and 1 mol of C 6 H 8 O 7, which is the limiting reactant?

4. Chapter 9 Section 3 Limiting Reactants and Percentage Yield

5. Learning targets 1.Describe a method for determining which of two reactants is a limiting reactant. 2.Calculate the amount in moles or mass in grams of a product given the amounts of two reactants, one of which is in excess. 3.Distinguish between theoretical yield, actual yield and percent yield. 4.Calculate the percent yield.

6. Why is this important? You’d think that it’s a good idea to always have the perfect amount of each reactant present, but that’s not always the case. For example, if you have one reactant that is considerably more expensive or rare than the other, you’d want to make sure there was plenty of the other reactant to make sure that all of it reacts.

7. Limiting Reactants and Excess Reactants The limiting reactant is the one that, well, limits the amount of the other reactant that can combine and the amount of product that can form. (It’s the one you run out of first. Like graham crackers… or chocolate.) The substance that isn’t used up completely in a reaction is called the excess reactant. (Marshmallows played this part in our lab.)

8. Quick side note… A limiting reactant may also be referred to as a limiting reagent. It’s the same thing. Maybe it’s the classic… or traditional, conventional name. But like most things, I’m going with what’s in your book so you don’t get confused. Or more confused… yikes.

9. C(s) + O 2 (g)  CO 2 (g) One mole of carbon reacts with one mole of oxygen to form one mole of carbon dioxide. Suppose you could mix 5 mol C with 10 mol O 2 and allow the reaction to take place.

10. We’ve got more O 2 than we need! Carbon is the limiting reactant because it limits the amount of CO 2 that is formed. Oxygen is the excess reactant, and 5 mol O 2 will be left over at the end of the reaction.

11. Example.F: Silicon dioxide (quartz) is usually quite unreactive but reacts readily with hydrogen fluoride according to the following equation: SiO 2 + 4HF  SiF 4 + 2H 2 O If 6.0 mol HF are added to 4.5 mol SiO 2, which is the limiting reactant? We need a plan!

12. SiO 2 + 4HF  SiF 4 + 2H 2 O If 6.0 mol HF are added to 4.5 mol SiO 2, which is the limiting reactant? 1.Pick one of the products, in this case SiF 4. Use the given amounts of each reactant and calculate the amount of SiF 4 that could be produced. (This means 2 calculations) 2.Compare the amounts of SiF 4. 3.The LR is the one that produces the smallest number of moles of SiF 4. This is also the maximum amount of product that can be made.

13. SiO 2 + 4HF  SiF 4 + 2H 2 O If 6.0 mol HF are added to 4.5 mol SiO 2, which is the limiting reactant? The solution: 6.0 mol HF x 1 mol SiF 4 = 1.5 mol SiF 4 (produced) 4 mol HF 4.5 mol SiO 2 x 1 mol SiF 4 = 4.5 mol SiF 4 (produced) 1 mol SiO 2

14. Under ideal conditions, 6.0 mol HF makes 1.5 mol SiF 4 and 4.5 mol SiO 2 present makes 4.5 mol SiF 4. Because 1.5 mol is smaller than 4.5 mol, the HF is the limiting reactant and SiO 2 is the excess reactant.

15. Let’s try these. We’ll do #1 together because it’s a little different.

16. Sample.G The black oxide of iron, Fe 3 O 4, occurs in nature as the mineral magnetite. This substance can also be made in the lab by the reaction between red-hot iron and steam according to this equation: 3Fe (s) + 4H 2 O (g)  Fe 3 O 4 (s) + 4H 2 (g) a.When 36.0 g H 2 O are mixed with 67.0 g Fe, which is the limiting reactant? b.What mass in grams of black iron oxide is produced? c.What mass in grams of excess reactant remains when the reaction is completed?

17. 3Fe (s) + 4H 2 O (g)  Fe 3 O 4 (s) + 4H 2 (g) Let’s start with a). Convert g  moles. Calculate the moles of one of the products. Since the question asks for the mass of Fe 3 O 4, let’s pick it. 67.0 g Fe x1 mol Fe x 1 mol Fe 3 O 4 = 0.400 mol Fe 3 O 4 55.85 g Fe3 mol Fe 36.0 g H 2 O x 1 mol H 2 O x 1 mol Fe 3 O 4 = 0.499 mol Fe 3 O 4 18.02 g 4 mol H 2 O

18. b) What mass in grams of black iron oxide is produced? 0.400 mol Fe 3 O 4 x 231.55 g Fe 3 O 4 = 92.6 g Fe 3 O 4 1 mol Fe 3 O 4

19. c) What mass in grams of excess reactant remains when the reaction is completed? 0.400 mol Fe 3 O 4 x 4 mol H 2 O x 18.02 g H 2 O 1 mol Fe 3 O 4 1 mol H 2 O =28.8 g H 2 O consumed 36.0 g H 2 O – 28.8 g H 2 O consumed = 7.2 g H 2 O remaining ** This can only be calculated as the difference between the original amount and the reacting amount.

20. Let’s practice these.

21. Percent Yield… or how to measure the screw ups in lab. Let’s say that, like everybody else in the world, you screw stuff up on a more-or-less continuous basis. This isn’t to say that you’re a bad person – it’s just that you’re human, and human beings make mistakes all the time through either little or huge screw ups.

22. Types of error Human error. This category is varied and extremely creative. It can be partially accounted for by being extremely careful, but given that nobody is perfect, it will always exist to some extent or another.

23. Three Mile Island, 1979

24. Types of error Systematic error: The errors that are made the same way every time due to quirks or limitations in the procedure. Instrumental error: This type of error is pretty uncommon with the stuff we use in high school chemistry, given the simplicity of the equipment we use. But in a bigger lab – some machine can give you bad answers. (minimalized by calibrating regularly)

25. A very well-hidden version of human error Unknown error: You finish lab and you got a lousy answer. You were really careful, so it wasn’t human error. You made sure the procedure was wonderful and that the equipment was working properly. Here’s the deal: something messed up along the line somewhere. (Maybe lab partner is still spitting and hasn’t been telling you. Maybe the reagents were contaminated…)

26. Unfortunately… Our lovely stoichiometric calculations assume that we’re perfect in every way. If the calculation says that we’ll make 75.0 grams of a compound, then we’ll make 75.0 grams of a compound. It doesn’t take into account the inherent limitations to how well we can do things in the lab, and that these limitations result in smaller than expected quantities.

27. Fortunately… We have a way of figuring out how badly we messed up. It’s called percent yield. The percent yield of a chemical reaction is a comparison of the amount of product you actually made versus how much product your stoichiometry calculations predicted you’d make.

28. Here are the tidy definitions. Actual yield is what happens in the lab. It’s the measured amount of a product that you obtain from the real-deal reaction. Theoretical yield is from your stoichiometric calculation predicting the amount.

29. To find percent yield of a reaction: Percent Yield = actual yield x 100 theoretical yield Chemists working in industry need to calculate this efficiency when they’re deciding about a process’s profitability.

30. An example You do the calculation and predict you’ll make 75.0 grams of a compound and you actually make 58.0 grams of that compound in the lab. Percent yield = 58.0 grams X 100% = 77.3% 75.0 grams

31. Higher percent yield = fewer mistakes happened If you got a 10% yield, it means that 90% of your product is lost somewhere, never to be seen again. On the other hand, if you have a 90% yield, you’ve only lost 10% of your product, which is pretty good.

32. Exceptions to the rule If your answer is exactly 100.0%... it is suspicious at best. It implies that you’ve made up your data. It’s impossible to be perfect! If your answer is greater than 100%, either you’ve violated the law of conservation of mass or more likely, you have lots of impurities in your compound. **Keep these in mind in lab.

33. Example.H Chlorobenzene, C 6 H 5 Cl, is used in the production of many important chemicals, like aspirin, dyes and disinfectants. One method of preparing C 6 H 5 Cl is to react benzene, C 6 H 6, with chlorine: C 6 H 6 + Cl 2  C 6 H 5 Cl + HCl When 36.8 g C 6 H 6 react with an excess of Cl 2, the actual yield of C 6 H 5 Cl is 38.8 g. What is the percent yield of C 6 H 5 Cl ?

34. Solution: 1.First find the theoretical yield with your super stoichiometric powers. 36.8 g C 6 H 6 x 1 mol C 6 H 6 x 1 mol C 6 H 5 Cl x 112.56 g C 6 H 5 Cl 78.12 g 1 mol C 6 H 6 1 mol C 6 H 5 Cl = 53.0 g C 6 H 5 Cl

35. Then calculate the percent yield: Percent yield = 38.8 g X 100 = 73.2% 53.0 g Remember sig figs match the measured amount which is 36.8 g C 6 H 6.

36. Let’s work a few of these. They are in your notes.

37. And that, my friends, is Chapter 9 all packed up!

39. Next up: The lab! We need some data from a reaction to “see” the next section in action.

40. Homework: FINISH ALL THE PROBLEMS ON LIMITING REACTANTS.