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Presentation on theme: "Chemical Reactions: Predicting Products and Balancing This PowerPoint presentation will be FAR more effective if you view it as a slideshow This PowerPoint."— Presentation transcript:

1 Chemical Reactions: Predicting Products and Balancing This PowerPoint presentation will be FAR more effective if you view it as a slideshow This PowerPoint presentation will be FAR more effective if you view it as a slideshow Press shift-F5 to enter slideshow mode Press shift-F5 to enter slideshow mode Press spacebar to advance through the slides Press spacebar to advance through the slides Press backspace to go backwards though the slides Press backspace to go backwards though the slides Press escape to exit slideshow mode Press escape to exit slideshow mode 1

2 2 Predicting Reactions – Double Replacement Step 1: Write ONE of each ion directly above (or below) the reactants Step 2: Write ONE of each ion on the product side, but swap the anions Step 3: Write neutral compounds for the products Step 4: Balance atoms (or ions) using coefficients

3 AlCl 3 + MgSO 4 Al +3 Cl -1 Mg +2 SO 4 2-  Step 1: Step 2: Cl -1  Al 3+ Mg 2+ SO 4 2- + Step 3: Predicting Reactions – Double Replacement AlCl 3 + MgSO 4 Al +3 Cl -1 Mg +2 SO 4 2- AlCl 3 + MgSO 4 Cl -1  Al 3+ Mg 2+ SO 4 2- +Al +3 Cl -1 Mg +2 SO 4 2- + MgCl 2 Al 2 (SO 4 ) 3 Write the balanced chemical reaction for the reaction of aluminum chloride and magnesium sulfate.

4 Predicting Reactions – Double Replacement 2 AlCl 3(aq) + 3 MgSO 4(aq)  Al 2 (SO 4 ) 3(aq) + 3 MgCl 2(aq) Step 4: Al Cl Mg SO 4 Al Cl Mg SO 4 1 3 1 1 2 2 1 3 2 AlCl 3(aq) + 3 MgSO 4(aq)  Al 2 (SO 4 ) 3(aq) + 3 MgCl 2(aq) 2 6 1 1 2 6 3 3 2 6 3 3 initial 1 st 2 nd 3 rd BALANCED!!!

5 Ca +2 C 2 O 4 -2 Ni +3 C 2 H 3 O 2 -1  Step 1: Step 2: C 2 O 4 -2 Ca 2+ Ni 3+ C 2 H 3 O 2 -1 + Step 3: Predicting Reactions – Double Replacement CaC 2 O 4 + Ni(C 2 H 3 O 2 ) 3 + Ni 2 (C 2 O 4 ) 3 Ca(C 2 H 3 O 2 ) 2 Write the balanced chemical reaction for the reaction of calcium oxalate with nickel (III) acetate Ca +2 C 2 O 4 -2 Ni +3 C 2 H 3 O 2 -1  CaC 2 O 4 + Ni(C 2 H 3 O 2 ) 3 C 2 O 4 -2 Ca 2+ Ni 3+ C 2 H 3 O 2 -1 +Ca +2 C 2 O 4 -2 Ni +3 C 2 H 3 O 2 -1  CaC 2 O 4 + Ni(C 2 H 3 O 2 ) 3

6 Predicting Reactions – Double Replacement 3 CaC 2 O 4(s) + 2 Ni(C 2 H 3 O 2 ) 3(aq)  3 Ca(C 2 H 3 O 2 ) 2(aq) + 1 Ni 2 (C 2 O 4 ) 3(aq) Step 4: Ca C2O4C2O4 Ni C2H3O2C2H3O2 1 1 1 3 1 3 2 2 3 CaC 2 O 4(s) + 2 Ni(C 2 H 3 O 2 ) 3(aq)  3 Ca(C 2 H 3 O 2 ) 2(aq) + 1 Ni 2 (C 2 O 4 ) 3(aq) 3 3 1 3 3 3 2 6 3 3 2 6 initial 1 st 2 nd 3 rd Ca C2O4C2O4 Ni C2H3O2C2H3O2 BALANCED!!!

7 H+H+ Cl -1 Ti +4 OH -1  Step 1: Predicting Reactions – Double Replacement HCl (aq) + Ti(OH) 4 Write the balanced chemical equation for the reaction between hydrochloric acid and titanium (IV) hydroxide Step 2: Cl -1 H+H+ Ti 4+ OH -1 + Step 3:+ TiCl 4 HOH H+H+ Cl -1 Ti +4 OH -1  HCl (aq) + Ti(OH) 4 Cl -1 H+H+ Ti 4+ OH -1 +H+H+ Cl -1 Ti +4 OH -1  HCl (aq) + Ti(OH) 4

8 Predicting Reactions – Double Replacement 4 HCl (aq) + 1 Ti(OH) 4(s)  4 HOH (l) + 1 TiCl 4(aq) Step 4: H Cl Ti OH H Cl Ti OH 1 1 1 4 1 4 1 1 4 HCl (aq) + 1 Ti(OH) 4(s)  4 HOH (l) + 1 TiCl 4(aq) 4 4 1 4 4 4 1 4 initial 1 st 2 nd BALANCED!!!

9 Pb +4 NO 3 -1 Co +3 CrO 4 -2  Step 1: Step 2: NO 3 -1 Pb 4+ Co 3+ CrO 4 -2 + Step 3: Predicting Reactions – Double Replacement Pb(NO 3 ) 4 + Co 2 (CrO 4 ) 3 + Co(NO 3 ) 3 Pb(CrO 4 ) 2 Write the balanced chemical reaction for the reaction of lead (IV) nitrate with cobalt (III) chromate Pb +4 NO 3 -1 Co +3 CrO 4 -2  Pb(NO 3 ) 4 + Co 2 (CrO 4 ) 3 NO 3 -1 Pb 4+ Co 3+ CrO 4 -2 +Pb +4 NO 3 -1 Co +3 CrO 4 -2  Pb(NO 3 ) 4 + Co 2 (CrO 4 ) 3

10 3 Pb(NO 3 ) 4(aq) + 2 Co 2 (CrO 4 ) 3(s)  3 Pb(CrO 4 ) 2(s) + 2 Co(NO 3 ) 3(aq) Predicting Reactions – Double Replacement Step 4: Pb NO 3 Co CrO 4 1 4 2 3 1 3 1 2 3 Pb(NO 3 ) 4(aq) + 2 Co 2 (CrO 4 ) 3(s)  3 Pb(CrO 4 ) 2(s) + 2 Co(NO 3 ) 3(aq) 1 6 2 2 1 4 4 6 3 12 4 6 initial 1 st 2 nd 3 rd Pb NO 3 Co CrO 4 BALANCED!!! 3 12 4 6 4 th 3 6 2 6 2 nd 3 Pb(NO 3 ) 4(aq) + 2 Co 2 (CrO 4 ) 3(s)  3 Pb(CrO 4 ) 2(s) + 2 Co(NO 3 ) 3(aq) 4 3 Pb(NO 3 ) 4(aq) + 2 Co 2 (CrO 4 ) 3(s)  3 Pb(CrO 4 ) 2(s) + 4 Co(NO 3 ) 3(aq)

11 H+H+ SO 4 -2 Fe +3 OH -1  Step 1: Predicting Reactions – Double Replacement H 2 SO 4(aq) + Fe(OH) 3 Step 2: SO 4 -2 H+H+ Fe 3+ OH -1 + Step 3:+ Fe 2 (SO 4 ) 3 HOH H+H+ SO 4 -2 Fe +3 OH -1  H 2 SO 4(aq) + Fe(OH) 3 SO 4 -2 H+H+ Fe 3+ OH -1 +H+H+ SO 4 -2 Fe +3 OH -1  H 2 SO 4(aq) + Fe(OH) 3 Write the balanced chemical equation for the reaction between sulfuric acid and iron (III) hydroxide

12 Predicting Reactions – Double Replacement 3 H 2 SO 4(aq) + 2 Fe(OH) 3(s)  6 HOH (l) + 1 Fe 2 (SO 4 ) 3(aq) Step 4: H SO 4 Fe OH 2 1 1 3 1 3 2 1 3 H 2 SO 4(aq) + 2 Fe(OH) 3(s)  6 HOH (l) + 1 Fe 2 (SO 4 ) 3(aq) 2 1 2 6 6 3 2 6 6 3 2 6 initial 1 st 2 nd 3 rd H SO 4 Fe OH BALANCED!!!

13 H+H+ Br -1 Al +3 CO 3 -2  Step 1: Step 2: Br -1 H+H+ Al 3+ CO 3 2 + Step 3: Predicting and Balancing Reactions HBr (aq) + Al 2 (CO 3 ) 3(s) + AlBr 3 H 2 CO 3(aq) H+H+ Br -1 Al +3 CO 3 -2  HBr (aq) + Al 2 (CO 3 ) 3(s) Br -1 H+H+ Al 3+ CO 3 -2 +H+H+ Br -1 Al +3 CO 3 -2  HBr (aq) + Al 2 (CO 3 ) 3(s) Write the balanced chemical equation for the reaction between hydrobromic acid and solid aluminum carbonate

14 HBr (aq) + Al 2 (CO 3 ) 3(s)  H 2 CO 3(aq) + AlBr 3 When carbonic acid is formed, it immediately decomposes into carbon dioxide and water, so the equation above becomes: HBr (aq) + Al 2 (CO 3 ) 3(s)  CO 2(g) + H 2 O (l) + AlBr 3 Predicting and Balancing Reactions

15 3 HBr (aq) + 2 Al 2 (CO 3 ) 3(s)  3 CO 2(g) + 3 H 2 O (l) + 2 AlBr 3(aq) Step 4: H Br Al C 1 1 2 3 2 3 1 1 3 HBr (aq) + 2 Al 2 (CO 3 ) 3(s)  3 CO 2(g) + 3 H 2 O (l) + 2 AlBr 3(aq) 2 6 2 1 6 HBr (aq) + 2 Al 2 (CO 3 ) 3(s)  3 CO 2(g) + 3 H 2 O (l) + 2 AlBr 3(aq) 6 6 2 3 6 6 2 1 initial 1 st 2 nd 3 rd H Br Al C BALANCED!!! 6 6 2 3 4 th 6 HBr (aq) + 2 Al 2 (CO 3 ) 3(s)  3 CO 2(g) + 3 H 2 O (l) + 2 AlBr 3(aq) Predicting and Balancing Reactions O93959O3

16 Ti +4 PO 3 -3 Bi +5 SO 3 -2  Step 1: Step 2: PO 3 -3 Ti 4+ Bi 5+ SO 3 -2 + Step 3: Predicting Reactions – Double Replacement Ti 3 (PO 3 ) 4 + Bi 2 (SO 3 ) 5 + Bi 3 (PO 3 ) 5 Ti(SO 3 ) 2 Write the balanced chemical reaction for the reaction of titanium (IV) phosphite with bismuth (V) sulfite Ti +4 PO 3 -3 Bi +5 SO 3 -2  Ti 3 (PO 3 ) 4 + Bi 2 (SO 3 ) 5 PO 3 -3 Ti 4+ Bi 5+ SO 3 -2 +Ti +4 PO 3 -3 Bi +5 SO 3 -2  Ti 3 (PO 3 ) 4 + Bi 2 (SO 3 ) 5

17 5 Ti 3 (PO 3 ) 4(s) + 6 Bi 2 (SO 3 ) 5(aq)  15 Ti(SO 3 ) 2(aq) + 4 Bi 3 (PO 3 ) 5(s) Predicting Reactions – Double Replacement Step 4: Ti PO 3 Bi SO 3 3 4 2 5 1 5 3 2 5 Ti 3 (PO 3 ) 4(s) + 6 Bi 2 (SO 3 ) 5(aq)  15 Ti(SO 3 ) 2(aq) + 4 Bi 3 (PO 3 ) 5(s) 3 4 12 30 5 Ti 3 (PO 3 ) 4(s) + 6 Bi 2 (SO 3 ) 5(aq)  15 Ti(SO 3 ) 2(aq) + 4 Bi 3 (PO 3 ) 5(s) 1 20 12 2 5 Ti 3 (PO 3 ) 4(s) + 6 Bi 2 (SO 3 ) 5(aq)  15 Ti(SO 3 ) 2(aq) + 4 Bi 3 (PO 3 ) 5(s) 15 20 12 30 initial 1 st 3 rd Ti PO 3 Bi SO 3 BALANCED!!! 15 20 12 30 2 nd

18 Pb +2 ClO 3 -1 Cu + Br -1  Step 1: Predicting Reactions – Double Replacement Pb(ClO 3 ) 2(aq) + CuBr (aq) Step 2: ClO 3 -1 Pb +2 Cu + Br -1 + Step 3:+ CuClO 3 PbBr 2 Pb +2 ClO 3 -1 Cu + Br -1  Pb(ClO 3 ) 2(aq) + CuBr (aq) ClO 3 -1 Pb +2 Cu + Br -1 +Pb +2 ClO 3 -1 Cu + Br -1  Pb(ClO 3 ) 2(aq) + CuBr (aq) An aqueous solution of lead (II) chlorate is added to an aqueous solution of copper (I) bromide. What is the balanced equation for this reaction?

19 Predicting Reactions – Double Replacement 1 Pb(ClO 3 ) 2(aq) + 2 CuBr (aq)  1 PbBr 2(s) + 2 CuClO 3(aq) Step 4: Pb ClO 3 Cu Br Pb ClO 3 Cu Br 1 2 1 1 1 1 1 2 1 Pb(ClO 3 ) 2(aq) + 2 CuBr (aq)  1 PbBr 2(s) + 2 CuClO 3(aq) 1 2 2 2 1 2 2 2 initial 1 st 2 nd BALANCED!!!

20 Hg 2 +2 NO 3 -1 Au +3 I -1  Step 1: Step 2: NO 3 -1 Hg 2 +2 Au +3 I -1 + Step 3: Predicting Reactions – Double Replacement Hg 2 (NO 3 ) 2 + AuI 3 + Au(NO 3 ) 3 Hg 2 I 2 Write the balanced chemical reaction for the reaction of mercury (I) nitrate with gold (III) iodide Hg 2 +2 NO 3 -1 Au +3 I -1  Hg 2 (NO 3 ) 2 + AuI 3 NO 3 -1 Hg 2 +2 Au +3 I -1 +Hg 2 +2 NO 3 -1 Au +3 I -1  Hg 2 (NO 3 ) 2 + AuI 3

21 3 Hg 2 (NO 3 ) 2(aq) + 2 AuI 3(aq)  3 Hg 2 I 2(s) + 2 Au(NO 3 ) 3(aq) Predicting Reactions – Double Replacement Step 4: Hg 2 NO 3 Au I 1 2 1 3 1 3 1 2 3 Hg 2 (NO 3 ) 2(aq) + 2 AuI 3(aq)  3 Hg 2 I 2(s) + 2 Au(NO 3 ) 3(aq) 3 6 1 3 1 6 2 2 3 6 2 6 initial 1 st 3 rd Hg 2 NO 3 Au I BALANCED!!! 3 6 2 6 2 nd

22 Balancing Reactions 1 C 11 H 24 + 17 O 2(g)  11 CO 2(g) + 12 H 2 O (g) C H O 11 24 2 1 2 3 1 C 11 H 24 + 17 O 2(g)  11 CO 2(g) + 12 H 2 O (g) 11 2 23 1 C 11 H 24 + 17 O 2(g)  11 CO 2(g) + 12 H 2 O (g) 11 24 34 1 C 11 H 24 + 17 O 2(g)  11 CO 2(g) + 12 H 2 O (g) 11 24 34 initial 1 st 2 nd 3 rd C H O BALANCED!!!

23 Balancing Reactions 1 C 32 H 64 + 48 O 2(g)  32 CO 2(g) + 32 H 2 O (g) C H O 32 64 2 1 2 3 1 C 32 H 64 + 48 O 2(g)  32 CO 2(g) + 32 H 2 O (g) 32 2 65 1 C 32 H 64 + 48 O 2(g)  32 CO 2(g) + 32 H 2 O (g) 32 64 96 1 C 32 H 64 + 48 O 2(g)  32 CO 2(g) + 32 H 2 O (g) 32 64 96 initial 1 st 2 nd 3 rd C H O BALANCED!!!

24 Balancing Reactions 2 C 8 H 18 + 25 O 2(g)  8 CO 2(g) + 9 H 2 O (g) C H O 8 18 2 1 2 3 2 C 8 H 18 + 25 O 2(g)  8 CO 2(g) + 9 H 2 O (g) 8 2 17 2 C 8 H 18 + 25 O 2(g)  8 CO 2(g) + 9 H 2 O (g) 8 18 25 2 C 8 H 18 + 25 O 2(g)  8 CO 2(g) + 9 H 2 O (g) 16 36 2 initial 1 st 2 nd 3 rd C H O BALANCED!!! 16 36 50 4 th 2 C 8 H 18 + 25 O 2(g)  8 CO 2(g) + 9 H 2 O (g) 1618 16 36 50 3 rd 2 C 8 H 18 + 25 O 2(g)  16 CO 2(g) + 18 H 2 O (g)

25 Balancing Reactions 1 C 8 H 16 + 12 O 2(g)  8 CO 2(g) + 8 H 2 O (g) C H O 8 16 2 1 2 3 1 C 8 H 16 + 12 O 2(g)  8 CO 2(g) + 8 H 2 O (g) 8 2 17 1 C 8 H 16 + 12 O 2(g)  8 CO 2(g) + 8 H 2 O (g) 8 16 24 1 C 8 H 16 + 12 O 2(g)  8 CO 2(g) + 8 H 2 O (g) 8 16 24 initial 1 st 2 nd 3 rd C H O BALANCED!!!

26 Balancing Reactions 2 C 8 H 14 + 23 O 2(g)  8 CO 2(g) + 7 H 2 O (g) C H O 8 14 2 1 2 3 2 C 8 H 14 + 23 O 2(g)  8 CO 2(g) + 7 H 2 O (g) 8 2 17 2 C 8 H 14 + 25 O 2(g)  8 CO 2(g) + 7 H 2 O (g) 8 14 23 2 C 8 H 14 + 25 O 2(g)  8 CO 2(g) + 7 H 2 O (g) 16 28 2 initial 1 st 2 nd 3 rd C H O BALANCED!!! 16 28 46 4 th 2 C 8 H 14 + 23 O 2(g)  8 CO 2(g) + 7 H 2 O (g) 1614 16 28 46 3 rd 2 C 8 H 14 + 23 O 2(g)  16 CO 2(g) + 14 H 2 O (g)

27 Balancing Reactions 1 C 2 H 5 OH + 3 O 2(g)  2 CO 2(g) + 3 H 2 O (g) C H O 2 6 3 1 2 3 2 2 5 2 6 7 2 6 7 initial 1 st 2 nd 3 rd C H O BALANCED!!!

28 Balancing Reactions 2 C 6 H 5 CO 2 H + 15 O 2(g)  7 CO 2(g) + 3 H 2 O (g) C H O 7 6 4 1 2 3 7 2 15 2 C 6 H 5 CO 2 H + 15 O 2(g)  7 CO 2(g) + 3 H 2 O (g) 7 6 17 2 C 6 H 5 CO 2 H + 15 O 2(g)  7 CO 2(g) + 3 H 2 O (g) 14 12 6 initial 1 st 2 nd 3 rd C H O BALANCED!!! 16 28 34 4 th 2 C 6 H 5 CO 2 H + 15 O 2(g)  7 CO 2(g) + 3 H 2 O (g) 146 12 34 3 rd 2 C 6 H 5 CO 2 H + 15 O 2(g)  14 CO 2(g) + 6 H 2 O (g)

29 Ni +3 HCO 3 -1 Ni +3 CO 3 -2  Ni(HCO 3 ) 3(s) + H 2 O + CO 2 Ni 2 (CO 3 ) 3 Ni +3 HCO 3 -1  Ni(HCO 3 ) 3(s) CO 2 Ni +3 H2OH2O CO 3 -2 + Ni +3 HCO 3 -1  Ni(HCO 3 ) 3(s) Write the balanced chemical equation for the decomposition of nickel (III) hydrogen carbonate. Remember that all hydrogen carbonates produce carbonate, water, and carbon dioxide upon heating. H2OH2OCO 2 + +  + Predicting and Balancing Reactions – Decompositions

30 2 Ni(HCO 3 ) 3(s)  1 Ni 2 (CO 3 ) 3(s) + 3 H 2 O (g) + 3 CO 2(g) Ni C H O 1 3 3 9 2 4 2 12 2 Ni(HCO 3 ) 3(s)  1 Ni 2 (CO 3 ) 3(s) + 3 H 2 O (g) + 3 CO 2(g) 2 6 6 18 2 Ni(HCO 3 ) 3(s)  1 Ni 2 (CO 3 ) 3(s) + 3 H 2 O (g) + 3 CO 2(g) 2 6 2 16 2 Ni(HCO 3 ) 3(s)  1 Ni 2 (CO 3 ) 3(s) + 3 H 2 O (g) + 3 CO 2(g) 2 6 6 18 initial 1 st 2 nd 3 rd Ni C H O BALANCED!!!

31 Sn +4 HCO 3 -1 Sn +4 CO 3 -2  Sn(HCO 3 ) 4(s) + H 2 O + CO 2 Sn(CO 3 ) 2 Sn +4 HCO 3 -1  Sn(HCO 3 ) 4(s) CO 2 Sn +4 H2OH2O CO 3 -2 + Sn +4 HCO 3 -1  Sn(HCO 3 ) 4(s) Write the balanced chemical equation for the decomposition of tin (IV) hydrogen carbonate. Remember that all hydrogen carbonates produce carbonate, water, and carbon dioxide upon heating. H2OH2OCO 2 + +  + Predicting and Balancing Reactions – Decompositions

32 1 Sn(HCO 3 ) 4(s)  1 Sn(CO 3 ) 2(s) + 2 H 2 O (g) + 2 CO 2(g) Sn C H O 1 4 4 12 1 3 2 9 1 Sn(HCO 3 ) 4(s)  1 Sn(CO 3 ) 2(s) + 2 H 2 O (g) + 2 CO 2(g) 1 4 2 11 1 Sn(HCO 3 ) 4(s)  1 Sn(CO 3 ) 2(s) + 2 H 2 O (g) + 2 CO 2(g) 1 4 4 12 initial 1 st 2 nd Sn C H O BALANCED!!!

33 Li + HCO 3 -1 Li + CO 3 -2  LiHCO 3(s) + H 2 O + CO 2 Li 2 CO 3 Li + HCO 3 -1  LiHCO 3(s) CO 2 Li + H2OH2O CO 3 -2 + Li + HCO 3 -1  LiHCO 3(s) Write the balanced chemical equation for the decomposition of lithium hydrogen carbonate. Remember that all hydrogen carbonates produce carbonate, water, and carbon dioxide upon heating. H2OH2OCO 2 + +  + Predicting and Balancing Reactions – Decompositions

34 2 LiHCO 3(s)  1 Li 2 CO 3(s) + 1 H 2 O (g) + 1 CO 2(g) Li C H O 1 1 1 3 2 2 2 6 2 LiHCO 3(s)  1 Li 2 CO 3(s) + 1 H 2 O (g) + 1 CO 2(g) 2 2 2 6 initial 1 st Li C H O BALANCED!!!

35 Ba +2 HCO 3 -1 Ba +2 CO 3 -2  Ba(HCO 3 ) 2(s) + H 2 O + CO 2 BaCO 3 Ba +2 HCO 3 -1  Ba(HCO 3 ) 2(s) CO 2 Ba +2 H2OH2O CO 3 -2 + Ba +2 HCO 3 -1  Ba(HCO 3 ) 2(s) Write the balanced chemical equation for the decomposition of barium hydrogen carbonate. Remember that all hydrogen carbonates produce carbonate, water, and carbon dioxide upon heating. H2OH2OCO 2 + +  + Predicting and Balancing Reactions – Decompositions

36 1 Ba(HCO 3 ) 2(s)  1 BaCO 3(s) + 2 H 2 O (g) + 2 CO 2(g) Ba C H O 1 2 2 6 1 2 2 6 initial Ba C H O BALANCED!!!

37 Rb + CO 3 -2 Rb + O -2  Rb 2 CO 3(s) + CO 2 Rb 2 O Rb + CO 3 -2  Rb 2 CO 3(s) CO 2 Rb + O -2 + Rb + CO 3 -2  Rb 2 CO 3(s) Write the balanced chemical equation for the decomposition of rubidium carbonate. Remember that all carbonates produce oxide, and carbon dioxide upon heating. CO 2 +  Predicting and Balancing Reactions – Decompositions

38 1 Rb 2 CO 3(s)  1 Rb 2 O (s) + 1 CO 2(g) Rb C O 2 1 3 2 1 3 initial Rb C O BALANCED!!!

39 Cu 2+ CO 3 -2 Cu 2+ O -2  CuCO 3(s) + CO 2 CuO Cu 2+ CO 3 -2  CuCO 3(s) CO 2 Cu 2+ O -2 + Cu 2+ CO 3 -2  CuCO 3(s) Write the balanced chemical equation for the decomposition of copper (II) carbonate. Remember that all carbonates produce oxide, and carbon dioxide upon heating. CO 2 +  Predicting and Balancing Reactions – Decompositions

40 1 CuCO 3(s)  1 CuO (s) + 1 CO 2(g) Cu C O 1 1 3 1 1 3 initial Cu C O BALANCED!!!

41 Cr +3 CO 3 -2 Cr +3 O -2  Cr 2 (CO 3 ) 3(s) + CO 2 Cr 2 O 3 Cr +3 CO 3 -2  Cr 2 (CO 3 ) 3(s) CO 2 Cr +3 O -2 + Cr +3 CO 3 -2  Cr 2 (CO 3 ) 3(s) Write the balanced chemical equation for the decomposition of chromium (III) carbonate. Remember that all carbonates produce oxide, and carbon dioxide upon heating. CO 2 +  Predicting and Balancing Reactions – Decompositions

42 1 Cr 2 (CO 3 ) 3(s)  1 Cr 2 O 3(s) + 3 CO 2(g) Cr C O 2 3 9 2 1 5 1 Cr 2 (CO 3 ) 3(s)  1 Cr 2 O 3(s) + 3 CO 2(g) 2 3 9 initial 1 st Cr C O BALANCED!!!

43 Mn 4+ CO 3 -2 Mn 4+ O -2  Mn(CO 3 ) 2(s) + CO 2 MnO 2 Mn 4+ CO 3 -2  Mn(CO 3 ) 2(s) CO 2 Mn 4+ O -2 + Mn 4+ CO 3 -2  Mn(CO 3 ) 2(s) Write the balanced chemical equation for the decomposition of manganese (IV) carbonate. Remember that all carbonates produce oxide, and carbon dioxide upon heating. CO 2 +  Predicting and Balancing Reactions – Decompositions

44 1 Mn(CO 3 ) 2(s)  1 MnO 2(s) + 2 CO 2(g) Mn C O 1 2 6 1 1 4 1 Mn(CO 3 ) 2(s)  1 MnO 2(s) + 2 CO 2(g) 1 2 6 initial 1 st Mn C O BALANCED!!!


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