Complete and balance the following reactions. Turn in answers corresponding to the eighth digit of your Student ID Number AND the eighth digit ±5, whichever gives a single digit. In all cases the assume that the second reagent shown is in significant excess.
(0) AlCl3 + NaOH(aq)
(1) CdCl2 + NaCN(aq)
(2) CoCl2 + NaSCN(aq)
(3) MgCl2 + NaOH(aq)
(4) PbCl2 + NaOH(aq)
(5) AgC2H3O2 + HCl(aq)
(6) CaC2O4 + HCl(aq)
(7) CdC2O4 + HCl(aq)
(8) CuCN + HCl(aq)
(9) ZnC2O4 + HCl(aq)
(0) AlCl3(aq) + 4 NaOH(aq) → ← [Al(OH)4]–(aq) + 4 Na+(aq) + 4 Cl–(aq)
(1) CdCl2(aq) + 4 NaCN(aq) → ← [Cd(CN)4]2–(aq) + 4 Na+(aq) + 2 Cl–(aq)
(2) CoCl2(aq) + 4 NaSCN(aq) → ← [Co(SCN)4]2–(aq) + 4 Na+(aq) + 2 Cl–(aq)
(3) MgCl2(aq) + 2 NaOH(aq) → Mg(OH)2(s) + 2 Na+(aq) + 2 Cl–(aq)
(4) PbCl2(s) + 3 NaOH(aq) → ← [Pb(OH)3]–(aq) + 3 Na+(aq) + 2 Cl–(aq)
(5) AgC2H3O2(aq) + HCl(aq) → AgCl(s) + HC2H3O2(aq)
(6) CaC2O4(s) + 2 HCl(aq) → Ca2+(aq) + 2 Cl–(aq) + H2C2O4(aq)
(7) CdC2O4(s) + 2 HCl(aq) → Cd2+(aq) + 2 Cl–(aq) + H2C2O4(aq)
(8) CuCN(s) + HCl(aq) → CuCl(s) + HCN(aq)
(9) ZnC2O4(s) + 2 HCl(aq) → Zn2+(aq) + 2 Cl–(aq) + H2C2O4(aq)