Answer:
a. K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. NO REACTION
Explanation:
For the reactions, the cation and the anion of the compounds will be replaced. The reaction will occur if at least one of the products is insoluble and will form a precipitated.
a. Potassium carbonate = K₂CO₃
Lead(II) nitrate = Pb(NO₃)₂
Products = KNO₃ and PbCO₃.
According to the solubility rules, all K⁺ ions are soluble, with no exceptions, so KNO₃ is soluble. All CO₃⁻² ions are insoluble, and Pb⁺² is not an exception, so PbCO₃ will be insoluble and will form a precipitated, so the reaction happen:
K₂CO₃(aq) + Pb(NO₃)₂(aq) → 2KNO₃(aq) + PbCO₃(s)
b. Lithium sulfate = Li₂SO₄
Lead(II) acetate = Pb(C₂H₃O₂)₂
Products = Li(C₂H₃O₂) and PbSO₄
All Li⁺ are solubles, without exceptions, so Li(C₂H₃O₂) is soluble, and all SO₄⁻² are soluble, but Pb⁺² is an exception, so PbSO₄ is insoluble and will form a precipitated, then the reaction happens:
Li₂SO₄(aq) + Pb(C₂H₃O₂)₂(aq) → 2Li(C₂H₃O₂) + PbSO₄(s)
c. Copper(II) nitrate = Cu(NO₃)₂
Magnesium sulfide = MgS
Products = CuS and Mg(NO₃)₂
All NO₃⁻ are soluble, with no exceptions, so Mg(NO₃)₂ is soluble, and all S⁺² are insoluble, and Cu⁺² is not an exception, so CuS is insoluble, and will form a precipitated, then the reaction happens:
Cu(NO₃)₂(aq) + MgS(aq) → Mg(NO₃)₂(aq) + CuS(s)
d. Strontium nitrate = Sr(NO₃)₂
Potassium iodi = KI
Products = K(NO₃)₂ and SrI₂
All K⁺ are soluble, with no exceptions, so K(NO₃)₂ is soluble, and all I⁻ are soluble, and Sr⁺² are not an exception, then SrI₂ is soluble. Therefore, no precipitated is formed and the reaction doesn't happen.
