HW 7 KEY!! Chap. 7, #'s 11, 12, 15-21 odd, 31, 33, 35, 39, 40, 53, 59, 67, 70, 72-75 all, 77, 82, 84, 88, 89 (plus a couple of unassigned ones) 11) NOTE: I used the solubility rules that I have provided for you here, not the ones in the book. a) PbS insoluble in water (rule #6) b) Mg(OH) 2 insoluble in water (rule #6) c) Na 2 SO 4 soluble in water (rule #1) d) (NH 4 ) 2 S soluble in water (rule #1) e) BaCO 3 insoluble in water (rule #7) f) AlPO 4 insoluble in water (rule #7) g) PbCl 2 insoluble in water (exception to rule #3) h) CaSO 4 soluble in water (rule #5) 12 a) Ba(NO 3 ) 2 soluble in water (rule #2) b) K 2 SO 4 soluble in water (rule #1) c) PbSO 4 insoluble in water (rule #5) d) Cu(OH) 2 insoluble in water (rule #6) e) KCl soluble in water (rule #1) f) Hg 2 Cl 2 insoluble in water (rule #3) g) (NH 4 ) 2 CO 3 soluble in water (rule #1) h) Cr 2 S 3 insoluble in water (rule #6) 15 a) CuCl 2(aq) + (NH 4 ) 2 S (aq) à CuS (s) + NH 4 Cl (aq) rule #2 rule #1 rule #6 rule #1 b) Ba(NO 3 ) 2(aq) + K 3 PO 4(aq) à Ba 3 (PO 4 ) 2(s) + KNO 3(aq) rule #2 rule #1 rule #7 rule #1 c) AgC 2 H 3 O 2(aq) + CaCl 2(aq) à AgCl (s) + Ca(C 2 H 3 O 2 ) 2(aq) rule #2 rule #3 exception to rule #3 rule #2 d) K 2 CO 3(aq) + CoCl 2(aq) à KCl (aq) + CoCO 3(s) rule #1 rule #2 rule #1 rule #7 e) H 2 SO 4(aq) + Ca(NO 3 ) 2(aq) à HNO 3(aq) + CaSO 4(aq) acid rule #1 acid rule #5 f) Hg 2 (C 2 H 3 O 2 ) 2(aq) + HCl (aq) à Hg 2 Cl 2(s) + HC 2 H 3 O 2(aq) rule #2 acid exception to rule #3 acid 17 a) 2 NH 4 Cl (aq) + H 2 SO 4(aq) à (NH 4 ) 2 SO 4(aq) + 2 HCl (aq) no ppt because all are strong electrolytes b) 2 K 2 CO 3(aq) + SnCl 4(aq) à 4 KCl (aq) + Sn(CO 3 ) 2(s) c) 2 NH 4 Cl (aq) + Pb(NO 3 ) 2(aq) à 2 NH 4 NO 3(aq) + PbCl 2(s) d) CuSO 4(aq) + 2 KOH (aq) à Cu(OH) 2(s) + K 2 SO 4(aq) e) Na 3 PO 4(aq) + CrCl 3(aq) à 3 NaCl (aq) + CrPO 4(s) f) 3 (NH 4 ) 2 S (aq) + 2 FeCl 3(aq) à 6 NH 4 Cl (aq) + Fe 2 S 3(s) 19 a) Na 2 SO 4(aq) + CaCl 2(aq) à CaSO 4(s) + 2 NaCl (aq) (Note: The books rules say CaSO 4 is insoluble, just go with it) b) Co(C 2 H 3 O 2 ) 2(aq) + Na 2 S (aq) à CoS (s) + 2 NaC 2 H 3 O 2(aq) c) 2 KOH (aq) + NiCl 2(aq) à Ni(OH) 2(s) + 2 KCl (aq)
NH + 2 4 SO 4 Ba 2+ NO 3 NH + 4 NO 3 Ba 2+ 2 SO 4 21 a) (NH 4 ) 2 SO 4(aq) + Ba(NO 3 ) 2(aq) à 2 NH 4 NO 3(aq) + BaSO 4(s) BaSO 4 is the ppt H + S 2 Ni 2+ 2 SO 4 H + 2 SO 4 Ni 2+ S 2 b) H 2 S (aq) + NiSO 4(aq) à H 2 SO 4(aq) + NiS (s) NiS is the ppt Fe 3+ Cl Na + OH Fe 3+ OH Na + Cl c) FeCl 3(aq) + 3 NaOH (aq) à Fe(OH) 3(s) + 3 NaCl (aq) Fe(OH) 3 is the ppt 31) A strong acid is an acid that disassociates 100% in water. In other words, every molecule of a strong acid will split into H + ions and the remaining anion. For example, if you put HNO 3 into water, every molecule will break apart into hydrogen ions and nitrate ions. HNO 3(aq) à H + (aq) + NO 3 (aq) Because they form so many ions in water, the 7 strong acids are strong electrolytes. 33) For all STRONG acid with a base, the net ionic equation is H + (aq) + OH (aq) à HOH (l) This is NOT true of weak acids. 35) If 1000 NaOH units were dissolved in a sample of water, the NaOH would produce 1000 Na + ions and 1000 OH ions. 39 a) HCl (aq) + KOH (aq) à HOH (l) + KCl (aq) b) RbOH (aq) + HNO 3(aq) à RbNO 3(aq) + HOH (l) c) HClO 4(aq) + NaOH (aq) à HOH (l) + NaClO 4(aq) d) HBr (aq) + CsOH (aq) à HOH (l) + CsBr (aq) 40) Think about these as a double replacement reaction between the salt and water K + 2 SO 4 H + OH K + OH H + 2 SO 4 a) K 2 SO 4(aq) + 2 HOH (l) à 2 KOH (aq) + H 2 SO 4(aq) So the base must have been KOH and the acid H 2 SO 4(aq) b) NaNO 3(aq) + HOH (l) à NaOH (aq) + HNO 3(aq) c) CaCl 2(aq) + 2 HOH (l) à Ca(OH) 2(aq) + 2 HCl (aq) d) Ba(ClO 4 ) 2(aq) + 2 HOH (l) à Ba(OH) 2(aq) + 2 HClO 4(aq) 53 a) A solid (BaSO 4 ) is formed, so double replacement (precipitation) b) An element (Zn) is added to an acid (HCl (aq) ) or ionic compound, so single replacement (redox) c) A solid (AgCl) is formed, so double replacement (precipitation) d) There is an acid and a base, so acid-base (neutralization and double replacement) e) An element (Zn) is added to an acid or ionic compound (CaSO 4 ), so single replacement (redox) f) I hate this one, skip it. (This is TECHNICALLY a neutralization and double replacement) g) There is an acid and a base, so acid-base (neutralization and double replacement) h) An element (Mg) is added to an acid or ionic compound (ZnCl 2 ), so single replacement (redox) i) A solid (BaSO 4 ) is formed, so double replacement (precipitation) 59 a) 2 C 6 H 6(l) + 9 O 2(g) à 12 CO 2(g) + 6 HOH (g) b) C 5 H 12(l) + 8 O 2(g) à 5 CO 2(g) + 6 HOH (g) c) C 2 H 6 O (l) + 3 O 2(g) à 2 CO 2(g) + 3 HOH (g)
60 a) C 3 H 8(g) + 10 O 2(g) à 3 CO 2(g) + 4 HOH (g) b) C 2 H 4(g) + 3 O 2(g) à 2 CO 2(g) + 2 HOH (g) c) 2 C 8 H 18(l) + 25 O 2(g) à 16 CO 2(g) + 18 HOH (g) (typo in questions, there should not be water in the reactants) 67) The molecular equation shows all compounds as molecules or formula units no matter how they exist in water The ionic equation shows all compounds as they actually exist in water. Strong electrolytes exist as ions in water, so they are shown as ions in the ionic equation. Weak and nonelectrolytes exist as molecules/formula units in water so that is how they are shown in the ionic equation. The net ionic equation most clearly shows the species that actually react with each other because all of the spectator ions are removed between the IE and the NIE. 70 a) HNO 3(aq) + KOH (aq) à HOH (l) + KNO 3(aq) b) H 2 SO 4(aq) + Ba(OH) 2(aq) à 2 HOH (l) + BaSO 4(s) c) HClO 4(aq) + NaOH (aq) à HOH (l) + NaClO 4(aq) d) 2 HCl (aq) + Ca(OH) 2(aq) à 2 HOH (l) + CaCl 2(aq) 72 a) 2 AgNO 3(aq) + H 2 SO 4(aq) à Ag 2 SO 4(s) + 2 HNO 3(aq) b) Ca(NO 3 ) 2(aq) + H 2 SO 4(aq) à CaSO 4(s) + 2 HNO 3(aq) c) Pb(NO 3 ) 2(aq) + H 2 SO 4(aq) à PbSO 4(s) + 2 HNO 3(aq) 73 a) FeCl 3(aq) + 3 NaOH (aq) à Fe(OH) 3(s) + 3 NaCl (aq) rule #3 rule #1 rule #6 rule #1 b) Ni(NO 3 ) 2(aq) + (NH 4 ) 2 S (aq) à NiS (s) + 2 NH 4 NO 3(aq) rule #2 rule #1 rule #6 rule #1 c) AgNO 3(aq) + KCl (aq) à AgCl (s) + KNO 3(aq) rule #2 rule #1 exception to rule #3 rule #1 d) Na 2 CO 3(aq) + Ba(NO 3 ) 2(aq) à 2 NaNO 3(aq) + BaCO 3(s) rule #1 rule #2 rule #1 rule #7 e) 2 KCl (aq) + Hg 2 (NO 3 ) 2(aq) à 2 KNO 3(aq) + Hg 2 Cl 2(s) rule #1 rule #2 rule #1 exception to rule #3 f) H 2 SO 4(aq) + Ba(NO 3 ) 2(aq) à HNO 3(aq) + BaSO 4(aq) acid rule #1 acid rule #5 74 a) NaOH + H 2 SO 4(aq) à Na 2 SO 4(aq) + HOH (l) b) RbOH + HNO 3(aq) à RbNO 3(aq) + HOH (l) c) KOH + HClO 4(aq) à KClO 4(aq) + HOH (l) d) KOH + HCl (aq) à KCl (aq) + HOH (l)
75 a) ME: HCl (aq) + AgNO 3(aq) à AgCl (s) + HNO 3(aq) IE: H + (aq) + Cl (aq) + Ag + (aq) + NO 3 (aq) à AgCl (s) + H + (aq) + NO 3 (aq) H + (aq) + Cl (aq) + Ag + (aq) + NO 3 (aq) à AgCl (s) + H + (aq) + NO 3 (aq) NIE: Cl (aq) + Ag + (aq) à AgCl (s) b) ME: 3 CaCl 2(aq) + 2 Na 3 PO 4(aq) à Ca 3 (PO 4 ) 2(s) + 6 NaCl (aq) IE: 3 Ca +2 (aq) + 6 Cl (aq) + 6 Na + 3 (aq) + 2 PO 4 (aq) à Ca 3 (PO 4 ) 2(s) + 6 Na + (aq) + 6 Cl (aq) 3 Ca +2 (aq) + 6 Cl (aq) + 6 Na + 3 (aq) + 2 PO 4 (aq) à Ca 3 (PO 4 ) 2(s) + 6 Na + (aq) + 6 Cl (aq) NIE: 3 Ca +2 3 (aq) + 2 PO 4 (aq) à Ca 3 (PO 4 ) 2(s) c) ME: Pb(NO 3 ) 2(aq) + BaCl 2(aq) à PbCl 2(s) + Ba(NO 3 ) 2(aq) IE: Pb +2 (aq) + 2 NO 3 (aq) + Ba +2 (aq) + 2 Cl (aq) à PbCl 2(s) + Ba +2 (aq) + 2 NO 3 (aq) Pb +2 (aq) + 2 NO 3 (aq) + Ba +2 (aq) + 2 Cl (aq) à PbCl 2(s) + Ba +2 (aq) + 2 NO 3 (aq) NIE: Pb +2 (aq) + 2 Cl (aq) à PbCl 2(s) d) ME: FeCl 3(aq) + 3 NaOH (aq) à Fe(OH) 3(s) + 3 NaCl (aq) IE: Fe +3 (aq) + 3 Cl (aq) + 3 Na + (aq) + 3 OH (aq) à Fe(OH) 3(s) + 3 Na + (aq) + 6 Cl (aq) Fe +3 (aq) + 3 Cl (aq) + 3 Na + (aq) + 3 OH (aq) à Fe(OH) 3(s) + 3 Na + (aq) + 3 Cl (aq) NIE: Fe +3 (aq) + 3 OH (aq) à Fe(OH) 3(s) 77) Again, think about these as a double replacement reaction between the salt and water a) KClO 4(aq) + HOH (l) à KOH (aq) + HClO 4(aq) b) CsNO 3(aq) + HOH (l) à CsOH (aq) + HNO 3(aq) c) KCl (aq) + HOH (l) à KOH (aq) + HCl (aq) d) Na 2 SO 4(aq) + 2 HOH (l) à NaOH (aq) + H 2 SO 4(aq) 81 a) Particles changes number of electrons (H + à H o, Fe o à Fe +2 ), so oxidation reduction b) Acid (HClO 4 ) and base (RbOH) as reactants, so neutralization c) Particles changes number of electrons (O o à O 2, Ca o à Ca +2 ), so oxidation reduction d) Acid (H 2 SO 4 ) and base (NaOH), so neutralization e) Solid formed (PbCO 3 ), so precipitation (all particles retain charge, no acid or base) f) Solid formed (CaSO 4 ), so precipitation (all particles retain charge, no acid or base) g) Acid (HNO 3 ) and base (KOH), so neutralization h) Solid formed (NiS), so precipitation (all particles retain charge, no acid or base) i) Particles changes number of electrons (Ni o à Ni +2, Cl o à Cl ), so oxidation reduction 82 all three are true
84 a) 2 NaHCO 3(s) à Na 2 CO 3(s) + H 2 O (g) + CO 2(g) b) 2 NaClO 3(s) à 2 NaCl (s) + 3 O 2(g) c) 2 HgO (s) à 2 Hg (l) + O 2(g) d) C 12 H 22 O 11(s) à 12 C (s) + 11 H 2 O (g) e) 2 H 2 O 2(l) à 2 H 2 O (l) + O 2(g) 88 a) Na à Na + + 1 e b) K à K + + 1 e c) Mg à Mg +2 + 2 e d) Ba à Ba +2 + 2 e e) Al à Al +3 + 3 e 89 a) O + 2 e à O 2 b) F + 1 e à F c) N + 3 e à N 3 d) Cl + 1 e à Cl e) S + 2 e à S 2 91 a) decomposition, redox 2 I 4 O 9(s) à 2 I 2 O 6(s) + 2 I 2(s) + 3 O 2(g) b) single replacement, redox Mg (s) + 2 AgNO 3(aq) à Mg(NO 3 ) 2(aq) + 2 Ag (s) c) single replacement, redox 2 Mg (s) + SiCl 4(l) à 2 MgCl 2(aq) + Si (s) d) double replacement, precipitation CuCl 2(aq) + 2 AgNO 3(aq) à Cu(NO 3 ) 2(aq) + 2 AgCl (s) e) combination, redox 2 Al (s) + 3 Br 2(l) à 2 AlBr 3(s)