Q (S) + ML (AQ) L (S) + QM (AQ) Is this reaction exo or endothermic? 2Al 2 O 3(S) + heat 4Al (S) + 3O 2(G) CaCrO 4(AQ) + K (S)

Transcription

1 CH 4(G) + 2O 2(G) 2H 2 O (G) + CO 2(G) set up C 5 H 12 + O 2 PQ P + Q 2Mg (S) + O 2(G) 2MgO (S) + heat H 2 O 2 O 2 + H 2 O PF 5(S) P (S) + F 2(G) RZ (AQ) + QP (AQ) RP (AQ) + QZ (S) Fe 2 S 3(S) + heat 2Fe (S) + 3S (S) set up N 2 + 3H 2 2NH 3 Be(ClO 4 ) 2(AQ) + MgCrO 4(AQ) WX + Y WXY N 2(G) + O 2(G) + heat 2NO (G) set up set up Li 3 PO 4(AQ) + Ca(HCO 3 ) 2(AQ) NaCl (AQ) + Li (S) Q (S) + ML (AQ) L (S) + QM (AQ) CH 4(G) + 2O 2(G) 2H 2 O (G) + CO 2(G) set up 2Al 2 O 3(S) + heat 4Al (S) + 3O 2(G) CaCrO 4(AQ) + K (S) 2Mg (S) + O 2(G) 2MgO (S) 2H 2(G) + O 2(G) 2H 2 O (G)

2 Mg (S) + O 2(G) 2MgO (S) + heat Energy is a product = exothermic PQ P + Q This represents decomposition C 5 H 12 + O 2 combustion CH 4(G) + 2O 2(G) 2H 2 O (G) + CO 2(G) This is combustion Fe 2 S 3(S) + heat 2Fe (S) + 3S (S) RZ (AQ) + QP (AQ) RP (AQ) + QZ (S) PF 5(S) P (S) + F 2(G) H 2 O 2 O 2 + H 2 O Energy is a reactant = endothermic This represents double replacement This is decomposition This is decomposition N 2(G) + O 2(G) + heat 2NO (G) Energy is a reactant = endothermic WX + Y WXY This represents synthesis Be(ClO 4 ) 2(AQ) + MgCrO 4(AQ) double replacement N 2 + 3H 2 2NH 3 This is synthesis CH 4(G) + 2O 2(G) 2H 2 O (G) + CO 2(G) Combustion is ALWAYS exothermic Q (S) + ML (AQ) L (S) + QM (AQ) This represents single replacement NaCl (AQ) + Li (S) single replacement Li 3 PO 4(AQ) + Ca(HCO 3 ) 2(AQ) double replacement 2Al 2 O 3(S) + heat 4Al (S) + 3O 2(G) Energy is a reactant = endothermic 2H 2(G) + O 2(G) 2H 2 O (G) This is synthesis 2Mg (S) + O 2(G) 2MgO (S) This is synthesis CaCrO 4(AQ) + K (S) single replacement

3 .. N 2(G) + O 2(G) NO (G) 2Nb (S) + Cl 2(G) 2NbCl 5(S) RZ (AQ) + QP (AQ) RP (AQ) + QZ (AQ) 3Li 2 S (AQ) + 2AlCl 3(AQ) Ca (S) + N 2(G) Ca 3 N 2(S) Mn (S) + 3Cl 2(G) 2MnCl 3(S) NaCl (AQ) + AgNO 3(AQ) Ba(NO 3 ) 2(AQ) + 2KOH (AQ) 2Be (S) + O 2(G) BeO (S) Br 2(L) + 2NH 4 I (AQ) NH 4 Br (AQ) + I 2(S) (NH 4 ) 2 S (AQ) + CaCl 2(AQ) NaHCO 3(AQ) + CaCrO 4(AQ) Fe 2 S 3 2Fe + S F 2(G) + 2NaBr (AQ) Br 2(L) + NaF (AQ) NaClO 3(AQ) + Al 2 (CO 3 ) 3(AQ) 2Na 3 PO 4(AQ) + 3Be(NO 3 ) 2(AQ) 2Mg + O 2 2MgO Zn (S) + HCl (AQ) ZnCl 2(AQ) + H 2(G) 2NH 4 Br (AQ) + Pb(NO 3 ) 2(AQ) 2NaOH (AQ) + Mg(C 2 H 3 O 2 ) 2(AQ)

4 3Li 2 S (AQ) + 2AlCl 3(AQ) Al 2 S 3 will be solid = YES RZ (AQ) + QP (AQ) RP (AQ) + QZ (AQ) These are symbols, not real compounds. 2 AQ products = no reaction 2Nb (S) + 5Cl 2(G) 2NbCl 5(S) N 2(G) + O 2(G) 2NO (G) Ba(NO 3 ) 2(AQ) + 2KOH (AQ) NaCl (AQ) + AgNO 3(AQ) Both products will be AQ, so NO, no double replacement here AgCl will be solid = YES 2Mn (S) + 3Cl 2(G) 2MnCl 3(S) 3Ca (S) + N 2(G) Ca 3 N 2(S) NaHCO 3(AQ) + CaCrO 4(AQ) Both products will be AQ, so NO, no double replacement here (NH 4 ) 2 S (AQ) + CaCl 2(AQ) CaS will be solid = YES Br 2 + 2NH 4 I 2NH 4 Br + I 2 Phase symbols omitted for space 2Be (S) + O 2(G) 2BeO (S) 2Na 3 PO 4(AQ) + 3Be(NO 3 ) 2(AQ) NaClO 3(AQ) + Al 2 (CO 3 ) 3(AQ) Be 3 (PO 4 ) 2(S) forms = YES Both products will be AQ, so NO, no double replacement here F 2 + 2NaBr Br 2 + 2NaF Phase symbols omitted for space Fe 2 S 3 2Fe + 3S 2NaOH (AQ) + Mg(C 2 H 3 O 2 ) 2(AQ) Mg(OH) 2(S) forms = YES 2NH 4 Br (AQ) + Pb(NO 3 ) 2(AQ) PbBr 2(S) forms = YES Zn + 2HCl ZnCl 2 + H 2 Phase symbols omitted for space 2Mg + 1O 2 2MgO We don t write one but it is 1.

5 Cs 2 S Mg 3 (PO 4 ) 2 Zn (S) + 2HCl (AQ) ZnCl 2(AQ) + H 2(G) F 2(L) + 2NH 4 I (AQ) 2NH 4 F (AQ) + I 2(S) Use table J, is this reaction possible? SrI 2 K 2 CO 3 F 2(G) + 2NaBr (AQ) Br 2(L) + 2NaF (AQ) Cl 2(L) + 2NH 4 I (AQ) Use table J, is this reaction possible? CuSO 4 NaOH Br 2(L) + 2NH 4 I (AQ) 2NH 4 Br (AQ) + I 2(S) Rb (S) + Ni(ClO 3 ) 2(AQ) Use table J, is this reaction possible? TiCl 4 Ba(C 2 H 3 O 2 ) 2 Ba(ClO 3 ) 2(AQ) + 2Li (S) 2LiClO 3(AQ) + Ba (S) Ni (S) + Rb(ClO 3 ) 2(AQ) CaCl 2 NH 4 C 2 H 3 O 2 3Al (S) + 2Ni(ClO 3 ) 2(AQ) 3Al(ClO 3 ) 3(AQ) + 2Ni (S) Use table J, is this reaction possible? Co (S) + Fe(NO 3 ) 2(AQ)

6 ANION replacement F 2(L) + 2NH 4 I (AQ) 2NH 4 F (AQ) + I 2(S) CATION replacement Zn (S) + 2HCl (AQ) ZnCl 2(AQ) + H 2(G) Mg 3 (PO 4 ) 2(AQ) Cs 2 S (S) Cl 2(L) + 2NH 4 I (AQ) Yes, NH 4 Cl AQ) forms ANION replacement F 2(G) + 2NaBr (AQ) Br 2(L) + 2NaF (AQ) K 2 CO 3(AQ) SrI 2(AQ) Rb (S) + Ni(ClO 3 ) 2(AQ) Yes, RbClO 3 forms ANION replacement Br 2(L) + 2NH 4 I (AQ) 2NH 4 Br (AQ) + I 2(S) NaOH (AQ) CuSO 4(S) Ni (S) + Rb(ClO 3 ) 2(AQ) No, Rb is higher on table J than Ni CATION replacement Ba(ClO 3 ) 2(AQ) + 2Li (S) 2LiClO 3(AQ) + Ba (S) Ba(C 2 H 3 O 2 ) 2(AQ) TiCl 4(AQ) Co (S) + Fe(NO 3 ) 2(AQ) No, Fe is higher on table J than Co CATION replacement 3Al (S) + 2Ni(ClO 3 ) 2(AQ) 3Al(ClO 3 ) 3(AQ) + 2Ni (S) NH 4 C 2 H 3 O 2(AQ) CaCl 2(AQ)

7 Use table F, is this double replacement? CuNO 3(AQ) + CaCl 2(AQ) HNO 3(AQ) Ba (S) + SnCl 4(AQ) F 2(G) + 2HCl (AQ) Use table F, is this double replacement? AgHCO 3(AQ) + FeBr 2(AQ) NH 3(AQ) Li (S) + KCl (AQ) Ca (S) + HCl (AQ) Use table F, is this double replacement? CaCrO 4(AQ) + Na 2 S (AQ) HCl (AQ) Zn (S) + CuSO 4(AQ) Ti (S) + SnBr 2(AQ) Use table F, is this double replacement? (NH 4 ) 3 PO 4(AQ) + RbNO 3(AQ) KOH (AQ) Ag (S) + AuF 3(AQ) Al (S) + Ba(ClO 3 ) 2(AQ) H 3 PO 4(AQ) H 2 SO 4(AQ) Mg (S) + NaCl (AQ) Sr (S) + Pb(NO 4 ) 4(AQ)

8 F 2(G) + 2HCl (AQ) fluorine replaces chlorine here. Ba (S) + SnCl 4(AQ) barium replaces tin here. HNO 3(AQ) This is nitric acid CuNO 3(AQ) + CaCl 2(AQ) Yes, double replacement here CuCl 2(S) forms Ca (S) + HCl (AQ) Li (S) + KCl (AQ) NH 3(AQ) AgHCO 3(AQ) + FeBr 2(AQ) calcium replaces hydrogen here. lithium replaces potassium here. This is ammonia, which is a weak base No double replacement here, both products are AQ Ti (S) + SnBr 2(AQ) titanium replaces tin here. Zn (S) + CuSO 4(AQ) zinc replaces copper here. HCl (AQ) This is hydrochloric acid CaCrO 4(AQ) + Na 2 S (AQ) Yes, double replacement here CaS (S) forms Al (S) + Ba(ClO 3 ) 2(AQ) Ag (S) + AuF 3(AQ) KOH (AQ) (NH 4 ) 3 PO 4(AQ) + RbNO 3(AQ) Single Replacement = No Al cannot replace Ba here. silver replaces gold here. This is potassium hydroxide, which is a strong base No double replacement here, both products are AQ Sr (S) + Pb(NO 4 ) 4(AQ) strontium replaces lead here. Mg (S) + NaCl (AQ) Single Replacement = No Mg cannot replace Na here. H 2 SO 4(AQ) This is sulfuric acid H 3 PO 4(AQ) This is Phosphoric acid