1. Lesson 11Stoichiometry Anything in black letters = write it in your notes (‘knowts’) 11.1 – How Much?? 11.2 – Mole Ratios 11.3 – Excess and Limiting Reactants 11.4 – Percent Yield 11.5 – Molarity

2. 11.1 – How Much?? Quantitative study of chemical rxns. In this lesson 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. 2  ( ) + 1  ( )  2  ( ) 6.02  10 23 molecules H 2 6.02  10 23 molecules O 2 6.02  10 23 molecules H 3 O 2 H 2 molecules + 1 O 2 molecule  2 H 2 O 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… 2 mol of H 2 + 1 mol of O 2 → 2 mol of H 2 O

5. The bike example… For simplicity, say a bike requires only 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. WheelsFrames Number of bikes that could be made How many wheels and/or frames leftover? 44 15 520 2 mol Complete each row in the chart…

9. MgO2O2 MgO Produced Excess Reactants 2 mol1 mol 2 mol 0.24 mol 0.036 mol 2.0 grams Complete each row in the chart using the given amounts 2Mg + O 2  2MgO

10. 11.2 – Mole Ratios A mole ratio comes from the coefficients of a balanced chemical equation. Mole ratios are used to compare the amount of mol of one substance to another.

11. 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…

12. N 2 (g) + 3H 2 (g)  2NH 3 (g) 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 How many moles of NH 3 are produced when 1.0 mol of N 2 reacts with excess H 2 ?

13. N 2 (g) + 3H 2 (g)  2NH 3 (g) Calculate the number of grams of NH 3 produced by the reaction of 5.40 g of hydrogen with an excess of nitrogen. = 31 g NH 3

14. 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 ? = 403 g P

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 = 918 g frames

17. The substance that is completely used up in a chemical rxn is called the limiting reactant. The substance that is NOT completely used up (and partially remains) is the excess reactant. 11.3 – Excess and Limiting Reactants

18. 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 1.26 mol Cu 1. Convert given amounts into moles. 2. Multiply either amount by the mole ratio. = 0.75 mol S 24.0 g S 2 mol Cu 1 mol S  = 1.26 mol Cu = 0.630 mol Cu 3. The smaller number of mol is the limiting reactant. Cu is the limiting reagent

19. 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.

20. 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% 11.4 – Percent Yield

21. CS 2 + 3Cl 2  CCl 4 + S 2 Cl 2 You Try It! What is the percent yield of CCl 4 if 617 g is produced from the reaction of 312 g of CS 2 ?

22. Practice QuizQuiz Practice Learn from mistakes Repeat Check answer key “Studying” is a myth Best way to study is to re-do practice quiz Use class time for work time and help Some points can be recovered in 1:1 time with Tischer

23. Quarter Grades 50% Tests/Quizzes 35% Classwork/Homework 15% Labs Semester Grades 45% 1 st Qtr 45% 2 nd Qtr 10% Semester Test Which of these matters the most??

24. Molarity – unit of solution concentration 11.5 – Molarity 1 Liter = 1000 mL

25. Calculate the molarity of these solutions. a) 0.55 mol NaOH dissolved in 1.0 L solution b) 4.0 grams of NaOH dissolved in 1.0 L solution c) 4.0 grams of NaOH dissolved in 250 mL solution

26. How many moles of solute would be in the following solutions? a) 1.00 Liter of 2.2 M HNO 3 b) 25.0 mL of 0.225 M HCl

27. To make 1.0 L of a 1.0 M NaOH solution, 1.0 mol NaOH = 40.0 grams NaOH