1. An insoluble precipitate or, as in an acid/base reaction, the
formation of water.
2. a. CaSO4 and Al(NO3)3
b. CaSO4 is insoluble
c. 3Ca(NO3)2(aq) + Al2(SO4)3(aq) → 3CaSO4(s) + 2Al(NO3)3(aq)
3. Use the solubility rules to check the solubility of the potential
products.
4. If the concentration of the ions are not great enough to make a
saturated solution, then there will be no precipitate.
5. 2KOH(aq) + Cu(NO3)2(aq) → Cu(OH)2(s) + 2KNO3(aq)
6. Cu2+ + 2OH- → Cu(OH)2(s)
7. You could mix one chemical that is both soluble in water and contains strontium with a second chemical that is water soluble and contains
the sulfate ion. Strontium nitrate and sodium sulfate are two possible compounds. In addition to insoluble strontium sulfate, SrSO4, a
solution of soluble NaCl is also produced, meaning that the solution contains sodium ions and chloride ions, Na+ and Cl-. But to avoid any
other salt in solution, you could use an acid and a base. The products of an acid-base reaction are a salt and water, and in this case the salt
is strontium sulfate. H2SO4(aq) + Sr(OH)2(aq) → SrSO4(s) + H2O(l)
8. 3Cu(NO3)2(aq) +2Na3PO4(aq) → Cu3(PO4)2(s) + 6NaNO3(aq) … and … 3Cu2+ + 2PO4- → Cu3(PO4)2(s)
Part 2 …. We did the first three reactions in class, and 8 and 9 are on the C7Chemistry website in the What’s Happening section for Tuesday.
4. AlCl3(aq) + Zn(C2H3O2)2(aq) → no reaction
5. 3K2CO3(aq) + 2NiCl3(aq) → Ni2(CO3)3(s) + 6KCl(aq)
2Ni3+ + 3CO32- → Ni2(CO3)3(s)
6. 2KI(aq) + Hg2(NO3)2(aq) → Hg2I2(s) + 2KNO3(aq)
Hg22+ + 2I- → Hg2I2(s)
7. 2Na3PO4(aq) + 3SnCl2(aq) → Sn3(PO4)2(s) + 6NaCl(aq)
3Sn2+ + 2PO43- → Sn3(PO4)2(s)
10. Pb(NO3)2(aq) + Na2S(aq) → PbS(s) + 2NaNO3(aq)
Pb2+ + S2- → PbS(s)
11. CuBr2(aq) + Fe(C2H3O2)3(aq) → no reaction
12. Bi(NO3)3(aq) + K3PO4(aq) → BiPO4(s) + 3KNO3(aq)
Bi3+ + PO43- → BiPO4(s)
13. AuCl3(aq) + 3LiOH(aq) → Au(OH)3(s) + 3LiCl(aq)
Au3+ + 3OH- → Au(OH)3(s)
14. FeBr2(aq) + Al2(SO4)3(aq) → no reaction
15. (NH4)2S (aq) + ZnCl2(aq) → ZnS(s) + 2NH4Cl(aq)
Zn2+ + S2- → ZnS(s)