2. 1 CHEMICAL REACTIONS Chapter 4 Reactants: Zn(s) + I 2 (s) Product: ZnI 2 (s)

3. 2 Chemical Equations Depict the kind of reactants and products and their relative amounts in a reaction. 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) The numbers in the front are called stoichiometric coefficients The letters (s), (g), and (l) are the physical states of compounds.

4. 3 Chemical Equations 4 Al(s) + 3 O 2 (g) ---> 2 Al 2 O 3 (s) This equation means 4 Al atoms + 3 O 2 molecules ---give---> 2 molecules of Al 2 O 3 2 molecules of Al 2 O 3 4 moles of Al + 3 moles of O 2 ---give---> 2 moles of Al 2 O 3 2 moles of Al 2 O 3

5. 4 Chemical Equations Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change.Because the same atoms are present in a reaction at the beginning and at the end, the amount of matter in a system does not change. The Law of the Conservation of MatterThe Law of the Conservation of Matter Demo of conservation of matter, See Screen 4.3.

6. 5 Because of the principle of the conservation of matter, an equation must be balanced. It must have the same number of atoms of the same kind on both sides. Chemical Equations Lavoisier, 1788

7. 6 Balancing Equations ___ Al(s) + ___ Br 2 (liq) ---> ___ Al 2 Br 6 (s)

8. 7 Balancing Equations ____C 3 H 8 (g) + _____ O 2 (g) ----> _____CO 2 (g) + _____ H 2 O(g) ____B 4 H 10 (g) + _____ O 2 (g) ----> ___ B 2 O 3 (g) + _____ H 2 O(g)

9. 8 STOICHIOMETRYSTOICHIOMETRY - the study of the quantitative aspects of chemical reactions.

10. 9 STOICHIOMETRYSTOICHIOMETRY It rests on the principle of the conservation of matter. 2 Al(s) + 3 Br 2 (liq) ------> Al 2 Br 6 (s)

11. 10 PROBLEM: If 454 g of NH 4 NO 3 decomposes, how much N 2 O and H 2 O are formed? What is the theoretical yield of products? STEP 1 Write the balanced chemical equation NH 4 NO 3 ---> N 2 O + 2 H 2 O

12. 11 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 2 Convert mass reactant (454 g) --> moles STEP 3 Convert moles reactant (5.68 mol) --> moles product

13. 12 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 3 Convert moles reactant --> moles product Relate moles NH 4 NO 3 to moles product expected. 1 mol NH 4 NO 3 --> 2 mol H 2 O Express this relation as the STOICHIOMETRIC FACTOR.

14. 13 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O = 11.4 mol H 2 O produced STEP 3 Convert moles reactant (5.68 mol) --> moles product

15. 14 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 4 Convert moles product (11.4 mol) --> mass product Called the THEORETICAL YIELD ALWAYS FOLLOW THESE STEPS IN SOLVING STOICHIOMETRY PROBLEMS!

16. 15 GENERAL PLAN FOR STOICHIOMETRY CALCULATIONS Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

17. 16 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 5 How much N 2 O is formed? Total mass of reactants = total mass of products 454 g NH 4 NO 3 = ___ g N 2 O + 204 g H 2 O mass of N 2 O = 250. g

18. 17 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield If you isolated only 131 g of N 2 O, what is the percent yield? This compares the theoretical (250. g) and actual (131 g) yields.

19. 18 454 g of NH 4 NO 3 --> N 2 O + 2 H 2 O STEP 6 Calculate the percent yield

20. 19 PROBLEM: Using 5.00 g of H 2 O 2, what mass of O 2 and of H 2 O can be obtained? 2 H 2 O 2 (liq) ---> 2 H 2 O(g) + O 2 (g) Reaction is catalyzed by MnO 2 Step 1: moles of H 2 O 2 Step 2: use STOICHIOMETRIC FACTOR to calculate moles of O 2 Step 3: mass of O 2

21. 20 Reactions Involving a LIMITING REACTANT In a given reaction, there is not enough of one reagent to use up the other reagent completely.In a given reaction, there is not enough of one reagent to use up the other reagent completely. The reagent in short supply LIMITS the quantity of product that can be formed.The reagent in short supply LIMITS the quantity of product that can be formed.

22. 21 LIMITING REACTANTS eactants R eactantsProducts 2 NO(g) + O 2 (g) 2 NO 2 (g) Limiting reactant = ___________ Excess reactant = ____________

23. 22 LIMITING REACTANTS Demo of limiting reactants on Screen 4.7

24. 23 Rxn 1: Balloon inflates fully, some Zn left * More than enough Zn to use up the 0.100 mol HCl Rxn 2: Balloon inflates fully, no Zn left * Right amount of each (HCl and Zn) Rxn 3: Balloon does not inflate fully, no Zn left. * Not enough Zn to use up 0.100 mol HCl LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2 123 (See CD Screen 4.8)

25. 24 Rxn 1Rxn 2Rxn 3 Rxn 1Rxn 2Rxn 3 mass Zn (g)7.003.271.31 mol Zn0.1070.0500.020 mol HCl0.1000.1000.100 mol HCl/mol Zn0.93/12.00/15.00/1 Lim ReactantLR = HClno LRLR = Zn LIMITING REACTANTS React solid Zn with 0.100 mol HCl (aq) Zn + 2 HCl ---> ZnCl 2 + H 2

26. 25 Reaction to be Studied 2 Al + 3 Cl 2 ---> Al 2 Cl 6

27. 26 PROBLEM: Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Mass reactant Stoichiometric factor Moles reactant Moles product Mass product

28. 27 Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

29. 28 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Reactants must be in the mole ratio Step 1 of LR problem: compare actual mole ratio of reactants to theoretical mole ratio.

30. 29 Deciding on the Limiting Reactant If There is not enough Al to use up all the Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Lim reag = Al

31. 30 If There is not enough Cl 2 to use up all the Al 2 Al + 3 Cl 2 ---> Al 2 Cl 6 Lim reag = Cl 2 Deciding on the Limiting Reactant

32. 31 We have 5.40 g of Al and 8.10 g of Cl 2 Step 2 of LR problem: Calculate moles of each reactant

33. 32 Find mole ratio of reactants This would be 3/2, or 1.5/1, if reactants are present in the exact stoichiometric ratio. Limiting reagent is Cl 2 2 Al + 3 Cl 2 ---> Al 2 Cl 6

34. 33 Mix 5.40 g of Al with 8.10 g of Cl 2. What mass of Al 2 Cl 6 can form? Limiting reactant = Cl 2 Base all calcs. on Cl 2 Limiting reactant = Cl 2 Base all calcs. on Cl 2 moles Cl 2 moles Al 2 Cl 6 grams Cl 2 grams Al 2 Cl 6 2 Al + 3 Cl 2 ---> Al 2 Cl 6

35. 34 CALCULATIONS: calculate mass of Al 2 Cl 6 expected. CALCULATIONS: calculate mass of Al 2 Cl 6 expected. Step 1: Calculate moles of Al 2 Cl 6 expected based on LR. Step 2: Calculate mass of Al 2 Cl 6 expected based on LR.

36. 35 Cl 2 was the limiting reactant.Cl 2 was the limiting reactant. Therefore, Al was present in excess. But how much?Therefore, Al was present in excess. But how much? First find how much Al was required. First find how much Al was required. Then find how much Al is in excess.Then find how much Al is in excess. How much of which reactant will remain when reaction is complete?

37. 36 2 Al + 3 Cl 2 products 0.200 mol 0.114 mol = LR Calculating Excess Al Excess Al = Al available - Al required = 0.200 mol - 0.0760 mol = 0.124 mol Al in excess

38. 37 Determining the Formula of a Hydrocarbon by Combustion CCR, page 138

39. 38 Using Stoichiometry to Determine a Formula Burn 0.115 g of a hydrocarbon, C x H y, and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. Burn 0.115 g of a hydrocarbon, C x H y, and produce 0.379 g of CO 2 and 0.1035 g of H 2 O. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O What is the empirical formula of C x H y ? What is the empirical formula of C x H y ?

40. 39 Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O Puddle of C x H y 0.115 g 0.379 g CO 2 +O 2 0.1035 g H 2 O 1 H 2 O molecule forms for each 2 H atoms in C x H y 1 CO 2 molecule forms for each C atom in C x H y

41. 40 Using Stoichiometry to Determine a Formula First, recognize that all C in CO 2 and all H in H 2 O is from C x H y. 1. Calculate amount of C in CO 2 8.61 x 10 -3 mol CO 2 --> 8.61 x 10 -3 mol C 2. Calculate amount of H in H 2 O 5.744 x 10 -3 mol H 2 O -- >1.149 x 10 -2 mol H C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O

42. 41 Using Stoichiometry to Determine a Formula Now find ratio of mol H/mol C to find values of “x” and “y” in C x H y. 1.149 x 10 -2 mol H/ 8.61 x 10 -3 mol C = 1.33 mol H / 1.00 mol C = 4 mol H / 3 mol C Empirical formula = C 3 H 4 C x H y + some oxygen ---> 0.379 g CO 2 + 0.1035 g H 2 O