1. 15 th Amendment Signed (1870) Reading for Wednesday Reading for Wednesday Chapter 4: Sections 4-6 Chapter 4: Sections 4-6 HOMEWORK – DUE Monday 2/8/16 (will accept on 2/10) HOMEWORK – DUE Monday 2/8/16 (will accept on 2/10) BW 3.2 (Bookwork) CH 2 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 BW 3.2 (Bookwork) CH 2 #’s 41, 45, 48, 56-61 all, 68, 70, 72, 73, 78, 79, 85, 88, 95, 96, 104, 119, 124 WS 4 (Worksheet) (from course website) WS 4 (Worksheet) (from course website) HOMEWORK – DUE Wednesday 2/10/16 HOMEWORK – DUE Wednesday 2/10/16 BW 4.1 (Bookwork) CH 4 #’s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 BW 4.1 (Bookwork) CH 4 #’s 3, 7, 9, 16, 17-31 odd, 42, 44, 46, 56, 62, 64, 66, 70, 72, 149 WS 5 (Worksheet) (from course website) WS 5 (Worksheet) (from course website) EXAM MONDAY EXAM MONDAY Lab Today/Tomorrow Lab Today/Tomorrow EXP 3 EXP 3 Lab Monday/Tuesday Lab Monday/Tuesday Lecture in Lab Lecture in Lab

2. 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) theoretical yield = 19.66g PbCl 2(s) Pb(NO 3 ) 2 is the L.R. (0 left) 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) 158.35 g/mol331.2 g/mol238.03 g/mol278.1 g/mol

3. 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? (Hint: start with the given amount of the limiting reactant.) Now calculate how much of the other product(s) will be formed. 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) 158.35 g/mol331.2 g/mol238.03 g/mol278.1 g/mol

4. 14.71 g of chromium (III) chloride is added to 23.41 g of lead (II) nitrate. How much of all reactants and products are present after the reaction? 19.66 g 11.22 g reactantsproducts CrCl 3 = 7.24 g Pb(NO 3 ) 2 = 0 g (L.R.) Cr(NO 3 ) 3 = 11.22 g PbCl 2 = 19.66 g Mass must be the same before and after the reaction!! mass before reaction  14.71 + 23.41 = 38.12 g mass after reaction  11.22 + 19.66 = 30.88 g difference will be the mass of excess reactant left over = 7.24 g 2 CrCl 3(aq) + 3 Pb(NO 3 ) 2(aq)  2 Cr(NO 3 ) 3(aq) + 3 PbCl 2(s) 14.71 g CrCl 3 (STARTED) – 7.462 g CrCl 3 (USED) = 7.24 g CrCl 3 (LEFT)

5. Oxidation Numbers/States

6. Monatomic ions Some charges can be predicted based on group number Some charges can be predicted based on group number 2+ Be A Bit of Review

7. Variable Oxidation State metals Charge can NOT be predicted based on periodic table Charge can NOT be predicted based on periodic table ANYTHING IN HERE CAN BE A VOS METAL(ALMOST)

8. A Bit of Review Variable Oxidation State metals Charge can NOT be predicted based on periodic table Charge can NOT be predicted based on periodic table Can assume multiple ion charges Can assume multiple ion charges non VOS metal – potassium ion = K + VOS metal – lead ion = Pb 2+ or Pb 4+ The charge of VOS metals MUST be indicated in the name with the use of roman numerals The charge of VOS metals MUST be indicated in the name with the use of roman numerals Sn = tin () ionSn = tin () ion Sn 2+ = tin (II) ionSn 4+ = tin (IV) ion

9. (III) CO 3 2- Co 2 (CO 3 ) 3 -6 cobaltcarbonate CO 3 2- Co CO 3 2- Total cation charges + total anion charges = 0 2x+= 0 x = 3 = oxidation state of Co Co CO 3 2-

10. (IV) CrO 4 2- Pb(CrO 4 ) 2 -4 leadchromate(IV) CrO 4 2- Pb CrO 4 2- Total cation charges + total anion charges = 0 1x+= 0 x = 4 = oxidation state of Pb

11. O 2- MnO 4 - -8 O 2- Mn O 2- Total atom “A” charges + total atom “B” charges = charge 1x+= x = 7 = oxidation state of Mn O 2- Total cation charges + total anion charges = 0

12. O 2- Cr 2 O 7 -2 -14 O 2- Cr O 2- Total atom “A” charges + total atom “B” charges = charge 2x+= x = 6 = oxidation state of Cr O 2- Cr -2

13. Rules for Assigning Oxidation Numbers 1) The sum of the oxidation numbers will always equal the particle’s charge 2(Co) + 3(-2) = 0 Co 2 (CO 3 ) 3 MnO 4 - 1(Mn) + -8 = -1 P 2 O 7 -4 2(P) + -14 = -4 1(C) + 3(-2) = -2 CO 3 2- Co = +3 C = +4 Mn = +7 P = +5

14. Rules for Assigning Oxidation Numbers 1) The sum of the oxidation numbers will always equal the particle’s charge 2) The oxidation number for a neutral atom is always zero 2 K (s) + 2 H 2 O (l)  2 KOH (aq) + H 2(g) Oxidation states are zero

15. Rules for Assigning Oxidation Numbers 1) The sum of the oxidation numbers will always equal the particle’s charge 2) The oxidation number for a neutral atom is always zero 3) Oxidation numbers for non-VOS metals depend on their group Li = +1Ba = +2Al = +3

16. Rules for Assigning Oxidation Numbers 1) The sum of the oxidation numbers will always equal the particle’s charge 2) The oxidation number for a neutral atom is always zero 3) Oxidation numbers for non-VOS metals depend on their group 4) Oxidation numbers for VOS metals are found based on anion 5) Oxidation numbers for nonmetals are typically found based on their group P = -3Se = -2Cl = -1

17. Rules for Assigning Oxidation Numbers 5) Oxidation numbers for nonmetals are typically found based on their group typically?!?! Fluorine = -1 always! Hydrogen = +1 unless paired with a metal Oxygen = -2 unless peroxide NaH (Na = +1, H = -1) BaO 2 (Ba = +2, O = -1) Other nonmetals: the element closest to fluorine on the PT gets to keep its “usual” O.S. OF 2 (F = -1, O = +2) ClO 3 -1 SCl 4 O.S. oxygen = -2 O.S. chlorine = +5O.S. sulfur = +4 O.S. chlorine = -1 P2S5P2S5 O.S. phosphorus = +5 O.S. sulfur = -2

18. O 2- Fe 3 O 4 -8 O 2- Fe O 2- Total atom “A” charges + total atom “B” charges = charge 3x+= x = 8/3 = oxidation state of Fe O 2- 0 Fe HUH??

19. O 2- Fe 3 O 4 -8 Fe O 2- Total atom “A” charges + total atom “B” charges = charge 3x+= x = 8/3 = oxidation state of Fe O 2- 0 Fe HUH??

20. Fe 3 O 4 Total atom “A” charges + total atom “B” charges = charge x = 8/3 = oxidation state of Fe Fe 3 O 4 = FeO + Fe 2 O 3 O.S. Fe = +2 O.S. Fe = +3 Average = 8/3

21. Practice!! Assign oxidation numbers to all atoms in each of the following: a) H 2 CO b) S 2 O 3 2- c) NH 4 + d) NO 3 - e) Br 2 f) Ca 3 (PO 4 ) 2 g) 2 K (s) + 2 H 2 O (l)  2 KOH (aq) + H 2(g)