1. Chapter 12 - Stoichiometry “SUPER DIMENSIONAL ANALYSIS”

2. Review of moles….. 1 mole = 6.022 x 10 23 particles Molar mass = X grams 1 mole X = molar mass of substance ( ) Use atomic mass

3. Calculate molar mass of CaBr 2 Ca: 40.1 x 1= 40.1 g/mol Br: 79.9 x 2 = 159.8 g/mol = 199.9 g/mol

4. What is mass of 0.250 mol CO 2 ? C: 12 x 1= 12 g/mol O:16 x 2 = 32 g/mol = 44 g/mol 0.250 mol CO 2 1 mol CO 2 44 g CO 2 = 11 g CO 2

5. Stoichiometry Defn – study of mass relationships between reactants and products in a chemical reaction What does this mean? –How much of something (products) can be made? –How much of starting materials were there?

6. Stoichiometry You must have a _____________________ to do stoichiometry calculations. balanced chemical reaction

7. To make a basic cheeseburger: 2 buns + 1 meat patty + 1 cheese 1 cheeseburger 1 c-burger 2 buns 1 c-burger 1 meat patty 1 c-burger 1 cheese 1 meat patty 2 buns 1 cheese 1 meat patty 2 buns 1 cheese

8. Mole Ratio Defn – ratio btwn # of moles of any two substances in a balanced chem rxn

9. Ex reaction 2 Al + 3 Br 2  2 AlBr 3 give all the mole ratios 2 mol Al 3 mol Br 2 2 mol Al 2 mol AlBr 3 2 mol Al 3 mol Br 2 2 mol AlBr 3 2 mol Al 2 mol AlBr 3 3 mol Br 2 2 mol AlBr 3

10. 4 steps to a basic stoichiometry problem 1)Balance the equation 2)Identify the given, convert to moles 3)Identify unknown, do a mole to mole ratio between given and unknown (KEY STEP) 4)Convert unknown to unit specified in problem

11. Stoichiometry Flow Chart grams Agrams Bmoles Amoles B Molar Mass A Molar Mass B Mole Ratio BABA

12. examples How many moles of H 2 are formed when 65 g HCl is used? If 3.7 mol KBr reacts with calcium, how many moles of CaBr 2 are formed? A A B B

13. Mole A to Mole B Conversions grams Bmoles Amoles B Molar Mass A Molar Mass B Mole Ratio BABA grams A

14. __ C 3 H 8 + __ O 2  __ CO 2 + __ H 2 O How many moles of CO 2 are produced when 10.0 moles of O 2 is used? 1453 10.0 mol ? mol 10.0 mol O 2 Mole Ratio BABA 5 mol O 2 3 mol CO 2 A B = 6 mol CO 2 Mole A to Mole B Conversions

15. __ C 3 H 8 + __ O 2  __ CO 2 + __ H 2 O How much O 2 was consumed to form 2.5 moles H 2 O? 1453 2.5 mol? mol 2.5 mol H 2 O Mole Ratio BABA 4 mol H 2 O 5 mol O 2 AB = 3.1 mol O 2

16. Mole A to Mass B Conversion grams Bmoles Amoles B Molar Mass A Molar Mass B Mole Ratio BABA grams A

17. ___ Na + ___ Cl 2  ___ NaCl How many grams of sodium chloride is formed when 1.25 moles of sodium react w/ chlorine gas? Mole A to Mass B Conversion 221 1.25 mol ? g 1.25 mol Na2 mol NaCl 2 mol Na1 mol NaCl 58.5 g NaCl A B Mole Ratio BABA Molar Mass B = 73.1 g NaCl

18. ___ Na + ___ Cl 2  ___ NaCl If 2.50 mol of NaCl is formed, how many grams of sodium was used? Mole A to Mass B Conversion 221 2.50 mol ? g 2.50 mol NaCl 2 mol NaCl 2 mol Na 1 mol Na 23 g Na AB Mole Ratio BABA Molar Mass B = 57.5 g Na

19. Mass A to Mole B Conversion grams Bmoles Amoles B Molar Mass A Molar Mass B Mole Ratio BABA grams A

20. ___ Na + ___ Cl 2  ___ NaCl How many moles of chlorine gas is needed to make 50 g NaCl? Mass A to Mole B Conversion 221 50 g ? mol 50 g NaCl 58.5 g NaCl 1 mol NaCl 2 mol NaCl 1 mol Cl 2 A B Mole Ratio BABA Molar Mass A = 0.43 mol Cl 2

21. Mass A to Mass B Conversion grams Bmoles Amoles B Molar Mass A Molar Mass B Mole Ratio BABA grams A

22. ___NH 4 NO 3  ___ N 2 O + ___ H 2 O Determine the mass of water formed from decomposition of 25.0 g NH 4 NO 3 Mass A to Mass B Conversion 1 12 25 g ? g 25 g NH 4 NO 3 80 g NH 4 NO 3 1 mol NH 4 NO 3 2 mol H 2 O 1 mol H 2 O 18 g H 2 O = 11.25 g H 2 O AB

23. Limiting Reactant In basic stoichiometry problems, you are provided with one given quantity. In LR problems, you are given both reactants. Before you can solve the problem, you have to determine which of the two given quantities to use as your given.

24. Limiting Reactant You need to choose the one that will run out first, known as the __________________. It controls how much product you can make. The other reactant is known as the _________________ because there will be some left over. limiting reactant excess reactant

25. 1 cheese + 2 bread slices  1 cheese sandwich Given: 10 pieces of cheese, 50 bread slices How many cheese sandwiches can you make? What is the limiting reactant? What is the excess reactant? How many pieces of excess reactant used? How many pieces of excess reactant left over? 10 20 30 cheese bread

26. How to find limiting reactant 1)Convert both amounts of reactants to moles 2)Divide the mole amount of each reactant by its coefficient in the balanced equation 3)Compare two numbers. The one that is smaller is the limiting reactant. Other one is excess reactant.

27. ____ Al + ___ CuCl 2  ___ Cu + ___ AlCl 3 2233 Find the limiting reactant if 6.9 g Al and 0.35 mol CuCl 2 are available. Reactant #1:Reactant #2: 6.9 g Al 1 mol Al 27 g Al = 0.256 mol Al = 0.35 mol CuCl 2 1) Convert both amounts of reactants to moles.

28. ____ Al + ___ CuCl 2  ___ Cu + ___ AlCl 3 2233 Reactant #1:Reactant #2: 0.256 mol Al0.35 mol CuCl 2 23 2) Divide each mole amount by its coefficient in the balanced equation = 0.128 = 0.117 CuCl 2 is the limiting reactant

29. ____ Al + ___ CuCl 2  ___ Cu + ___ AlCl 3 2233 Find the limiting reactant if 6.2 g Al and 48.5 g CuCl 2 are available. Reactant #1:Reactant #2: 6.2 g Al 1 mol Al 27 g Al = 0.230 mol Al = 0.360 mol CuCl 2 1) Convert both amounts of reactants to moles. 48.5 g CuCl 2 1 mol CuCl 2 134.5 g CuCl 2

30. ____ Al + ___ CuCl 2  ___ Cu + ___ AlCl 3 2233 Reactant #1:Reactant #2: 0.230 mol Al0.360 mol CuCl 2 23 2) Divide each mole amount by its coefficient in the balanced equation = 0.115 = 0.120 Al is the LR CuCl 2 is excess reactant

31. Based on the LR in #4, how many grams of copper will be produced? 1 mol Al 2 mol Al 3 mol Cu 1 mol Cu 63.5 g Cu ____ Al + ___ CuCl 2  ___ Cu + ___ AlCl 3 2233 = 21.9 g Cu 27 g Al 6.2 g Al

32. Percent Yield You buy a 500 g ketchup bottle, do you ever use all 500 g? No. Some is still left on the sides you cannot retrieve. Not all 100% is used.

33. Percent Yield Theoretical Yield – max amount of product that can be produced (what you expect to get) Actual Yield – amt of product you actually produced (always less than theoretical) Percent Yield = actual theoretical x 100

34. Ex problem #1 Joe does experiment to form carbon dioxide. The maximum he can obtain is 34.5 grams. He performs the experiment and only obtains 18.6 grams. What is his percent yield? ATAT = 18.6 g 34.5 g X 100 = 53.9%

35. Ex problem #2 What is the theoretical yield of Ag 2 CrO 4 if 0.500 g AgNO 3 reacts with excess K 2 CrO 4 ? __K 2 CrO 4 + __ AgNO 3  __KNO 3 + __Ag 2 CrO 4 2211 0.500 g ? g 0.5 g K 2 CrO 4 170 g K 2 CrO 4 1 mol K 2 CrO 4 1 mol Ag 2 CrO 4 331.7 g = 0.49 g Ag 2 CrO 4

36. Ex problem #2 If 0.455 g of Ag 2 CrO 4 is produced, what is the percent yield? ATAT = 0.455 g 0.49 g X 100 = 93.2%