1. 4-3 General Types (Sections 7.6, 7.7) And you

2. 1) Synthesis reaction elements come together to form solid (in general but exceptions occur) A + B → C or element + element → compound Use periodic table to determine charges of elements. Write product using criss-cross or balancing of charges to determine subscripts. Balance equation

3. Synthesis rxn. Demo Al (s) + I 2(s) → http://www.youtube.com/watch?v=8TXjxhuyc BU http://www.youtube.com/watch?v=8TXjxhuyc BU Now balance this puppy please 2 Al (s) + 3 I 2(s) → 2 AlI 3(g) You can see this is an example where a solid is not formed!!! AlI 3(g)

4. Synthesis Practice Na (s) + O 2(g) → K (s) + S 8(s) → Now balance them please Na 2 O (s) K 2 S (S) 2 4 8 16

5. 2) Decomposition reaction Solid heated to form solid and gas (in general) C → A + B or compound → element + element Reactants will consist of just one compound. Write compound’s elements as the products. Remember the 7 diatomics! Balance the equation

6. Decomposition Demo NI 3(s) → www.youtube.com/watch?v=DFfRqoIdArM Now please balance this puppy 2 NI 3(S)  N 2(g) + 3 I 2(g) N 2(g) + I 2(g)

7. Decomposition practice HgO (s) → Now balance it please 2HgO (s)  2Hg (l) + O 2 (g) H 2 O (l) → Products and balance please 2H 2 O (l)  2H 2 (g) + O 2 (g) Hg (l) + O 2 (g)

8. 3) Combustion reaction (my favorite) C x H y (or C x H y O z ) + O 2 → CO 2 + H 2 O Rapid burning; one of the reactants will be O 2. For hydrocarbons (compounds with C, H, and/or O), the products are:CO 2 + H 2 O

9. Combustion Demo CH 4 (g) + O 2 (g) → http://www.youtube.com/watch?v=q5W2trP0 PmU http://www.youtube.com/watch?v=q5W2trP0 PmU The Methane Mamba! Demo The Methane Mamba! Balance please CO 2 (g) + H 2 O (g)

10. Combustion Practice Recall the suggestion of starting with a 2 in front of the hydrocarbon, and reducing at the end if needed C 2 H 4 O 2 (l) + O 2 (g) → 2 C 2 H 4 O 2 (l) + 4 O 2 (g) → 4 CO 2 (g) + 4 H 2 O (g) C 2 H 4 O 2 (l) + 2 O 2 (g) → 2 CO 2 (g) + 2 H 2 O (g)

11. More combustion rxn. practice C 2 H 2 (g) + O 2 (g) → 2 C 2 H 2 (g) + 5 O 2 (g) → 4 CO 2 (g) + 2 H 2 O (g) Starting with the 2 really helps on this one!

12. 4) Single Replacement rxn. A + BC → AC + B or element + compound → compound + element Note: most take place in aqueous environment Recall the H.S. dance parallel Metals replace metals and nonmetals replace nonmetals. Replacement occurs only if the element is higher on the activity list than the element in the compound.

13. Activity list or metals Li K Ca Na Mg Al Zn Cr Fe Ni Sn (tin) Pb (lead) H Cu Hg Ag (silver) Pt Au (gold) So Li would replace K if K was in a compound Li is “more attractive” to that two-timing anion!!! I hate that Li !!!!!! Just kidding – don’t be a player hater

14. Activity list for non metals fluorine chlorine bromine Iodine (general trend – but Cl may be more active)

15. Write the new compound with balanced charges and the individual element (is it diatomic?????). If the element is lower on the list then the element in the compound, no reaction occurs. Kind of a weird chemical hierarchy thing

16. Single replacement rxn. demo Al (s) + CuCl 2(aq) → Demo in class 2 Al (s) + 3 CuCl 2(aq) → 3 Cu (s) + 2 AlCl 3 (aq) Note Al is higher = “more attractive” than Cu

17. Single replacement practice HCl (aq) + Zn (s) → 2 HCl (aq) + Zn (s) → ZnCl 2 (aq) + H 2 (g) MgCl 2(aq) + Zn (s) → MgCl 2(aq) + Zn (s) → No reaction !!!!! Cl 2(aq) + CaBr 2(aq) → Cl 2(aq) + CaBr 2(aq) → CaCl 2 + Br 2

18. 5) Double replacement AB(aq) + CD(aq) → AD + CB compound + compound → compound + compound Takes place in an aqueous environment Metal pairs with nonmetal from other compound and vice versa. Write new products with charges balancing. Consult solubility chart to determine the insoluble compound (ppt). This chart, seen below, is also found on the inside back cover of your text and on page 170.

19. 1. Most nitrate (NO 3 - ) salts are soluble. 2. Most salts of Na +, K +, and NH 4 + are soluble. 3. Most chloride salts are soluble. Notable exceptions are AgCl, PbCl 2, and Hg 2 Cl 2. 4. Most sulfate salts are soluble. Notable exceptions are BaSO 4, PbSO 4, and CaSO 4. 5. Most hydroxide (OH - ) compounds are only slightly soluble. The important exceptions are NaOH and KOH. Ba(OH) 2 and Ca(OH) 2 are only moderately soluble. 6. Most sulfide (S -2 ), carbonate (CO 3 -2 ), and phosphate (PO 4 -3 ) salts are only slightly soluble. Note: slightly soluble and moderately soluble are, essentially, the same as “insoluble”.

20. Posted on web page!!!!! (if you want your own copy)

21. Double replacement demo CuCl 2 (aq) + Na 2 CO 3 (aq) → http://www.youtube.com/watch?v=lEdAkBNN qSY http://www.youtube.com/watch?v=lEdAkBNN qSY CuCl 2 (aq) + Na 2 CO 3 (aq) → CuCO 3(s) + 2NaCl (aq) Note: see the solubility chart to know what forms if you do observe a ppt.

22. Practice double replacement rxns. NaOH (aq) + FeCl 3(aq) → 3NaOH (aq) + FeCl 3(aq) → 3NaCl (aq) + Fe(OH) 3 WOW that was FUN!!!!

23. Molecular, Complete Ionic, and Net Ionic Equations The full double replacement is also called the” molecular equation.” For example, the molecular equation from above is: 3NaOH (aq) + FeCl 3(aq) → 3NaCl (aq) + Fe(OH) 3(s)

24. Molecular, Complete Ionic, and Net Ionic Equations The full double replacement is also called the” molecular equation.” For example, the molecular equation from above is: 3NaOH (aq) + FeCl 3(aq) → 3NaCl (aq) + Fe(OH) 3(s) A variation of this equation is the “complete ionic equation”, where the (aq) compounds are broken into their respective ions: Like this:

25. 3NaOH (aq) + FeCl 3(aq) → 3NaCl (aq) + Fe(OH) 3(s) 3Na + (aq) + 3OH - (aq) + Fe +3 (aq) + 3Cl - (aq) → 3Na + (aq) + 3Cl - (aq) + Fe(OH) 3(s) Note how the solid remains as a compound. This equation more accurately depicts what is happening in solution, since (aq) means that the ions a surrounded by water molecules (“hydrated”):

27. A final version of the molecular equation is the “net ionic equation.” Simply put, this form only shows the ions that form the solid: 3Na + (aq) + 3OH - (aq) + Fe +3 (aq) + 3Cl - (aq) → 3Na + (aq) + 3Cl - (aq) + Fe(OH) 3(s) Cancel the redundant parts to find: 3OH - (aq) + Fe +3 (aq) → Fe(OH) 3(s)

28. Note:All equations, regardless of the variation, require the atoms and charges to balance and the phases of matter included Practice: Na 2 SO 4(aq) + BaCl 2(aq) → Na 2 SO 4(aq) + BaCl 2(aq) → 2NaCl (aq) + BaSO 4(s) Complete ionic: 2Na + + SO 4 -2 + Ba +2 + 2Cl -  2Na + + 2Cl - + Ba SO 4 (s) Net Ionic: SO 4 -2 (aq) + Ba +2 (aq)  BaSO 4 (S) NOW that was FUN!!!!