Question

Complete and balance the following methesis reaction in aqueous solution.  

Instructions
• Use proper element capitalization. Iron = Fe not fe.
• Enter formulas without subscrips: CaCl₂ = CaCl2, K₂SO₄ = K2SO4
• Monoatomic ions do not require parentheses. Correct: Na2CO3. Incorrect: (Na)2CO3
•Polyatomic ions only require parentheses when there are two or more. Correct: Na3PO4. Incorrect: Na3(PO4), Correct: NH4NO3. Incorrect: (NH4)(NO3). correct: Ca3(PO4)2  
• In each case one product is soluble (aq) and one insoluble (s). Use solubility rules to determine which product appears in each box.

_Pb(C₂H₃O₂)₂(aq) + _KI (aq) ➝__ (aq) + __(s)

Answer 1

Pb(C2H3O2)2 (aq) + 2 KI (aq) --------------------> PbI2 (s) + 2 C2H3O2K (aq)

here the lead acetate reacts with potassium iodide then forms the lead iodide which is insoluble and the potassium acetate are the products

and here the balanced equation is as follows

Pb(C2H3O2)2 (aq) + 2 KI (aq) --------------------> PbI2 (s) + 2 C2H3O2K (aq)

the precipitate is PbI2 and the all iodides are soluble except the Pb+ , Ag+ and the Hg22+