1. LECTURE TWELVE CHM 151 ©slg TOPICS: Overview: Balancing Limiting Reagent Precipitation Reactions Acid/Base Reactions

2. GROUP WORK, BALANCE C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O (sugar) C 5 H 11 OH + O 2 --> CO 2 + H 2 O (alcohol) a) Balance C b) Balance H c) Balance O d) Check

3. C 6 H 12 O 6 + O 2 --> CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + H 2 O C 6 H 12 O 6 + O 2 --> 6 CO 2 + 6 H 2 O C 6 H 12 O 6 + 6 O 2 --> 6 CO 2 + 6 H 2 O 6C 12H 6 O + 12 O ---> 6 C 12 O + 12 H 6 O Solution, Sugar 12 O + 6 O = 18 O

4. C 5 H 11 OH + O 2 --> CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + H 2 O C 5 H 11 OH + O 2 --> 5 CO 2 + 6 H 2 O C 5 H 11 OH + 7.5 or 15/2 O 2 --> 5 CO 2 + 6 H 2 O 1 O 15 O 10 O 6 O #1 #2 #3

5. C 5 H 11 OH + 7.5 or 15/2 O 2 --> 5 CO 2 + 6 H 2 O 10 O 6 O X 2 2 C 5 H 11 OH + 15 O 2 --> 10 CO 2 + 12 H 2 O [10C 24H 2 O] + 30 O --> [10C 20 O] + [24H 12 O]

6. Ch 4. Prob 24: Which is LR, what is theoretical yield, how much of excess reagent remains after reaction is complete? CO(g) + 2 H 2 (g) ----> CH 3 OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol 74.5 g 12.0 g ? g theoretical ? g excess reagent leftover Solving the question: How much of the excess reagent is left after reaction is theoretically complete?

7. CO(g) + 2 H 2 (g) ----> CH 3 OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol 74.5 g 12.0 g ? g theoretical 74.5 g CO 1 mol CO 1 mol CH 3 OH 32.04 g CH 3 OH 28.01 g CO 1 mol CO 1 mol CH 3 OH = 85.2 g CH 3 OH 12.0 g H 2 1 mol H 2 1 mol CH 3 OH 32.04 g CH 3 OH 2.106 g H 2 2 mol H 2 1 mol CH 3 OH = 95.4 g CH 3 OH TOO MUCH!

8. CO(g) + 2 H 2 (g) ----> CH 3 OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol 74.5 g ( 12.0 g)excess 85.2 g theoretical ? How much leftover Figure out how much is used up! Calculate from either value, LR or product from LR: 74.5 g CO 1 mol CO 2 mol H 2 2.016 g H 2 28.01 g CO 1 mol CO 1 mol H 2 = 10.7 g H 2 required 12.0 g H 2 (original) - 10.7 g H 2 (required) = 1.3 g H 2 (leftover)

9. CO(g) + 2 H 2 (g) ----> CH 3 OH (l) 28.01 g/mol 2.016 g/mol 32.04 g/mol 74.5 g ( 12.0 g)excess 85.2 g theoretical ? How much leftover 85.2 g CH 3 OH 1 mol CH 3 OH 2 mol H 2 2.016 g H 2 32.04 g CH 3 OH 1 mol CH 3 OH 1 mol H 2 = 10.7 g H 2 required 12.0 g H 2 (original) - 10.7 g H 2 (required) = 1.3 g H 2 (leftover)

10. Constant mass, Bromine, variable masses Fe Ch 4, #60

11. (a) What mass of Br 2 is used when the reaction consumes 2.0 g Fe? Product  10.8 g; 10.8 - 2.0 = 8.8 g Br (b) What is the mole ratio of Fe to Br in this reaction? 2.0 g Fe 1 mol Fe =.036 mol Fe.036 = 1 55.85 g Fe.036 8.8 g Br 1 mol Br =.11 mol Br.11 = 3 79.9 g Br.036

12. (c) What is the empirical formula of the product? FeBr 3 (d) Balanced chemical equation: Fe + 3 Br 2 -----> 2 FeBr 3 (e) Name? Iron (III) Bromide

13. Which best describes graph: 1. When 1.00 g Fe is added, Fe is LR 2. When 3.50 g of Fe is added, there is an excess of Br 2 3. When 2.50 g of Fe is added to the Br 2, both reactants are used up completely 4. When 2.00 g of Fe is added to the Br 2, 10.0 g of product is formed. The percent yield must be 20%

14. The Electrolyte is Usually Water Soluble if: THE CATION IS: Na + K + NH 4 + OR THE ANION IS*: Cl -, Br -, I - ClO 4 -, ClO 3 - NO 3 - SO 4 2- C 2 H 3 O 2 -

15. THE COMPOUND IS PROBABLY INSOLUBLE IF: THE CATION IS NOT: Na + K + NH 4 + THE ANION IS: CO 3 2- PO 4 3- S 2- O 2- OH -

16. There are a few notable exceptions to the solubility guide on the last slide, principally the ones noted below, which you should be aware of: Insoluble in Water: AgCl, AgBr, AgI PbCl 2, PbBr 2, PbI 2 BaSO 4, PbSO 4

17. Let’s consider names and formulas of common acids and bases which we meet in these reactions: COMMON STRONG ACIDS: HCl Hydrochloric acid HBr Hydrobromic acid HI Hydroiodic Acid HNO 3 Nitric Acid HClO 4 Perchloric Acid H 2 SO 4 Sulfuric Acid Note names: learn! Add H to anions: go from ide, ite, ate to “ic or “ous” acid”

18. H 2 O HCl (g) ---------> HCl (aq) -----> H + (aq) + Cl - (aq) “Hydrogen Chloride” “Hydrochloric Acid” HBr (g) “Hydrogen bromide” HBr (aq) “Hydrobromic acid” HI (g) “Hydrogen iodide” HI (aq) “Hydroiodic acid”

19. Nitrate, NO 3 - ---> HNO 3 Nitric Acid Nitrite, NO 2 - ---> HNO 2 Nitrous Acid Sulfate, SO 4 2- -----> H 2 SO 4 Sulfuric Acid Sulfite, SO 3 2- -----> H 2 SO 3 Sulfurous Acid Perchlorate, ClO 4 - -----> HClO 4 Perchloric Acid

20. COMMON WEAK ACIDS:* H 3 PO 4 Phosphoric Acid H 2 CO 3 Carbonic Acid H 2 SO 3 Sulfurous Acid CH 3 CO 2 H or HC 2 H 3 O 2 Acetic Acid * and many, many more..... Remember: These are not broken up in net ionic equations because they are “mostly molecular” in water...

21. Group Work: Do NIE’s for the following: Na 2 S (aq) + CrCl 3 (aq) ----> ? (NH 4 ) 3 PO 4 (aq) + Fe(CH 3 CO 2 ) 2 (aq) -----> ? 1. Write down ions, decide 2. Write out formulas of products 3. Write out equation and balance 4. Do total Ionic 5. Do Net Ionic

22. Group Work: Solutions Na 2 S (aq) + CrCl 3 (aq) ----> ? Step 1: Ions Involved: Na + (no ppt) S 2- (ppt???) Cr 3+ (ppt???) Cl - (no ppt) Step 2: Formulas of Products: Na + + Cl - ---> NaCl (aq) Cr 3+ + S 2- ----> Cr 2 S 3 (s)

23. Step 3: Write unbalanced equation: Na 2 S (aq) + CrCl 3 (aq) ----> NaCl (aq) + Cr 2 S 3 (s) Step 4: Balance: 3 Na 2 S (aq) + 2 CrCl 3 (aq) ----> 6 NaCl (aq) + Cr 2 S 3 (s) Step 5: Total Ionic Equation: 6 Na + (aq) + 3 S 2- (aq) + 2 Cr 3+ (aq) + 6 Cl - (aq) -----> 6 Na + (aq) + 6 Cl - (aq) + Cr 2 S 3 (s)

24. Step 5: Total Ionic Equation: 6 Na + (aq) + 3 S 2- (aq) + 2 Cr 3+ (aq) + 6 Cl - (aq) -----> 6 Na + (aq) + 6 Cl - (aq) + Cr 2 S 3 (s) Step 6: Net Ionic Equation: 2 Cr 3+ (aq) + 3 S 2- (aq) -----> Cr 2 S 3 (s)

25. (NH 4 ) 3 PO 4 (aq) + Fe(CH 3 CO 2 ) 2 (aq) -----> ? Step 1: Ions Involved: NH 4 + (no ppt) PO 4 3- (ppt???) Fe 2+ (ppt???) CH 3 CO 2 - (no ppt) Step 2: Formulas of Products: NH 4 + + CH 3 CO 2 - ---> NH 4 CH 3 CO 2 (aq) Fe 2+ + PO 4 3- ----> Fe 3 (PO 4 ) 2 (s)

26. Step 3: Write unbalanced equation: (NH 4 ) 3 PO 4 (aq) + Fe(CH 3 CO 2 ) 2 (aq) -----> NH 4 CH 3 CO 2 (aq) + Fe 3 (PO 4 ) 2 (s) Step 4: Balance: 2 (NH 4 ) 3 PO 4 (aq) + 3 Fe(CH 3 CO 2 ) 2 (aq) -----> 6 NH 4 CH 3 CO 2 (aq) + Fe 3 (PO 4 ) 2 (s)

27. 2 (NH 4 ) 3 PO 4 (aq) + 3 Fe(CH 3 CO 2 ) 2 (aq) -----> 6 NH 4 CH 3 CO 2 (aq) + Fe 3 (PO 4 ) 2 (s) Step 5: Total Ionic Equation: 6 NH 4 + (aq) + 2 PO 4 3- (aq) + 3 Fe 2+ (aq) + 6 CH 3 CO 2 - (aq) -----> 6 NH 4 + (aq) + 6 CH 3 CO 2 - (aq) + Fe 3 (PO 4 ) 2 (s) Step 6: Net Ionic Equation: 3 Fe 2+ (aq) + 2 PO 4 3- (aq) -----> Fe 3 (PO 4 ) 2 (s)

28. Group work 2: H 2 SO 4 (aq) + Al(OH) 3 (s) -----> ? H 3 PO 4 (aq) + KOH (aq) ----> HCH 3 CO 2 (aq) + Mg(OH) 2 (s) ----->? 1. Form H 2 O and use leftover ions to form salt 2. Write balanced equation 3. Do total ionic (watch out for weak acids!) 4. Do net ionic equation

29. Group Work 2 solutions: H 2 SO 4 (aq) + Al(OH) 3 (s) -----> ? ----> H 2 O + salt Step One: Formula of salt: Al 3+ + SO 4 2- -----> Al 2 (SO 4 ) 3 Step Two: Write Equation, Balance: H 2 SO 4 (aq) + Al(OH) 3 (s) -----> H 2 O + Al 2 (SO 4 ) 3 (aq) 3 H 2 SO 4 (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O + Al 2 (SO 4 ) 3 (aq)

30. Step 3: Total Ionic Equation: 6 H + (aq) + 3 SO 4 2- (aq) + 2 Al(OH) 3 (s) -----> 6 H 2 O + 2 Al 3+ (aq) + 3 SO 4 2- (aq) Step Four: Net Ionic Equation 6 H + (aq) + 6 OH - (aq) -----> 6 H 2 O (l) + 2 Al 3+ (aq)

31. H 3 PO 4 (aq) + KOH (aq) ----> ? -----> H 2 O + salt Step One: salt formula K + + PO 4 3- -----> K 3 PO 4 (aq) Step Two: Write Equation; Balance H 3 PO 4 (aq) + KOH (aq) ----> H 2 O (l) + K 3 PO 4 (aq) H 3 PO 4 (aq) + 3 KOH (aq) ---->3 H 2 O (l) + K 3 PO 4 (aq)

32. Step Three: Total Ionic: H 3 PO 4 (aq) + 3 K + (aq) + 3 OH - (aq) -----> 3 H 2 O (l) + 3 K + (aq) + PO 4 3- (aq) Step Four: Net Ionic: H 3 PO 4 (aq) + 3 OH - (aq) -----> 3 H 2 O (l) + PO 4 3- (aq)

33. HCH 3 CO 2 (aq) + Mg(OH) 2 (s) ----->? -----> H 2 O + salt Step One: Salt Formula: Mg 2+ + CH 3 CO 2 - -----> Mg(CH 3 CO 2 ) 2 (aq) Step Two: Write Equation, Balance: HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> H 2 O + Mg(CH 3 CO 2 ) 2 (aq) 2 HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> 2 H 2 O + Mg(CH 3 CO 2 ) 2 (aq)

34. Step Three: Total Ionic Equation: 2 HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> 2 H 2 O(l) + Mg 2+ (aq) +2 CH 3 CO 2 - (aq) Step Four: Net Ionic Equation 2 HCH 3 CO 2 (aq) + Mg(OH) 2 (s) -----> 2 H 2 O(l) + Mg 2+ (aq) +2 CH 3 CO 2 - (aq)