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Chemical Reactions Predicting Products. Combination (Synthesis) n the combination of 2 or more substances to form a single compound n only one product.

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Presentation on theme: "Chemical Reactions Predicting Products. Combination (Synthesis) n the combination of 2 or more substances to form a single compound n only one product."— Presentation transcript:

1 Chemical Reactions Predicting Products

2 Combination (Synthesis) n the combination of 2 or more substances to form a single compound n only one product A + B  AB

3 Combination (Synthesis) H 2 (g) + Cl 2 (g)  2 HCl(g)

4 Al (s) + Cl 2(g) AlCl 3(s) 2 3  2 Combination (Synthesis) n Products: n Single elements combine to form a compound  ionic - neutral (enough of each ion to cancel charges)  covalent - hard to tell – usually need more information given Metal oxides + water  metal hydroxide Nonmetal oxides + water  oxyacids

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6 Practice: Synthesis Reaction n Na + O 2 → n Mg + F 2 → Metal oxides + water  metal hydroxide n Li 2 O+ H 2 O → n CaO + H 2 O → n Al 2 O 3 + H 2 O →

7 Practice: Synthesis Reaction n Na + O 2 → Na 2 O n Mg + F 2 → Mg F 2 Metal oxides + water  metal hydroxide n Li 2 O+ H 2 O → LiOH n CaO + H 2 O →Ca(OH) 2 n Al 2 O 3 + H 2 O → Al(OH) 3

8 Decomposition n a compound breaks down into 2 or more simpler substances n only one reactant AB  A + B

9 Decomposition 2 H 2 O (l)  2 H 2(g) + O 2(g)

10 KBr (l)  2  2 K (s) + Br 2(l) Decomposition n Products:  binary - break into elements  others - hard to tell, need more info

11 Decomposition of oxyacids & bases Metal hydroxide  Metal oxides + water Sodium hydroxide… NaOH (aq)  2 Na 2 O (s) + H 2 O (l)

12 Practice Decomposition Metal hydroxide  Metal oxides + water n LiOH → n Ca(OH) 2 → n Al (OH) 3 → binary - break into elements n Na 2 O→ n MgF 2 →

13 Practice Decomposition Metal hydroxide  Metal oxides + water n LiOH → Li 2 O + H 2 O n Ca(OH) 2 → CaO + H 2 O n Al (OH) 3 → AlO 3 + H 2 O binary - break into elements n Na 2 O→ Na + O 2 n MgF 2 → Mg + F 2

14 Single Replacement n An element reacts with a compound n One element replaces another in a compound  metal replaces metal (+) OR  nonmetal replaces nonmetal (-) A + BC  AC + B

15 Single Replacement Cu (s) + 2AgNO 3 (aq)  Cu(NO 3 ) 2 (aq) + 2Ag (s)

16 Fe (s) + CuSO 4 (aq)  Cu(s)+ FeSO 4 (aq) Single Replacement n Products:  metal  metal cation (+) - (check activity series pg 333)  nonmetal  nonmetal anion (-) - (halogens = F more active than I)  free element must be more active or NR Cl 2 (l) + NaBr (aq)  NaCl + Br 2

17 Single Replacement Practice Halogens replace Halogens n Br 2 + CaI 2 -  Metals replace Metals n Al + CuCl 2 -  Metals replace Hydrogen n Mg + HCl -  n K + H 2 O -  n K + HOH 

18 Single Replacement Practice Halogens replace Halogens n Br 2 + CaI 2 -  Ca Br 2 + I 2 Metals replace Metals n Al + CuCl 2 -  Cu + Al 2 Cl 3 Metals replace Hydrogen n Mg + HCl -  H 2 + MgCl 2 n K + H 2 O -  KOH + H 2 n K + HOH  KOH + H 2

19 AB + CD  AD + CB Double Replacement n ions in two compounds “change partners” n cation of one compound combines with anion of the other

20 Double Replacement Pb(NO 3 ) 2(aq) + K 2 CrO 4 (aq)  PbCrO 4 (s) + 2KNO 3 (aq)

21 Pb(NO 3 ) 2(aq) + KI (aq)  PbI 2 (s) + KNO 3(aq) 2 2 Double Replacement Practice. NaOH (aq) + HCl (aq)  NaCl (aq) + H 2 O (l)

22 Double Replacement Practice Metals in an Ionic Compound will replace other metals in an ionic Compound n Al 2 (SO 4 ) 3 + Ca 3 (PO 4 ) 2 -  Cations will replace cations n Cd 3 (PO 4 ) 2 + (NH 4 ) 2 S -  Metals will replace Hydrogen n H 2 SO 4 + NaOH 

23 Double Replacement Practice Metals in an Ionic Compound will replace other metals in an ionic Compound n Al 2 (SO 4 ) 3 + Ca 3 (PO 4 ) 2 -  Ca(SO 4 ) + Al(PO 4 ) Cations will replace cations n Cd 3 (PO 4 ) 2 + (NH 4 ) 2 S -  (NH 4 ) 3 PO 4 + CdS Metals will replace Hydrogen n H 2 SO 4 + NaOH  HOH + Na 2 SO 4 n H 2 SO 4 + NaOH  H 2 O + Na 2 SO 4

24 Combustion CH 4(g) + 2O 2(g)  CO 2(g) + 2H 2 O (g) n the burning of any substance in O 2 to produce heat A + O 2  A x O y C x H y (O) + O 2  CO 2 + H 2 O

25 Na (s) + O 2 (g)  C 3 H 8(g) + O 2(g)  5 3 4 Combustion n Products:  contain oxygen  hydrocarbons form CO 2 + H 2 O CO 2 (g) + H 2 O (g) Na 2 O (s) 4 2

26 Practice Combustion n C 12 H 22 O 11 + 12O 2 -  n C 2 H 5 OC 2 H 5 + 6O 2 -  n Sr + O 2 -  n Al + O 2 ---> \

27 Practice Combustion n C 12 H 22 O 11 + 12O 2 ---> CO 2 + H 2 O n C 2 H 5 OC 2 H 5 + 6O 2 ---> CO 2 + H 2 O n Sr + O 2 ---> SrO n Al + O 2 ---> Al 2 O 3


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