1. Write the Balanced Chemical Equation

The reaction between copper(II) acetate (Cu(CH₃COO)₂) and sodium sulfide (Na₂S) produces copper(II) sulfide (CuS) precipitate and sodium acetate (NaCH₃COO) in solution.

The balanced equation is:

Cu(CH₃COO)₂(aq) + Na₂S(aq) → CuS(s) + 2 NaCH₃COO(aq)

2. Calculate Moles of Copper(II) Acetate

* Molarity = moles of solute / liters of solution

* We have 170 mL (0.170 L) of 0.30 M copper(II) acetate.

* Rearranging the formula: moles of solute = Molarity * Liters of solution

* moles of Cu(CH₃COO)₂ = 0.30 M * 0.170 L = 0.051 moles

3. Determine Limiting Reactant

* Since sodium sulfide is in excess, copper(II) acetate is the limiting reactant. This means the amount of precipitate formed will be determined by the amount of copper(II) acetate available.

4. Calculate Moles of Copper(II) Sulfide Precipitate

* The balanced equation shows a 1:1 mole ratio between Cu(CH₃COO)₂ and CuS.

* Therefore, 0.051 moles of Cu(CH₃COO)₂ will produce 0.051 moles of CuS.

5. Calculate Mass of Copper(II) Sulfide Precipitate

* Moles = Mass / Molar Mass

* The molar mass of CuS is 95.61 g/mol.

* Rearranging the formula: Mass = Moles * Molar Mass

* Mass of CuS = 0.051 moles * 95.61 g/mol = 4.88 grams

Therefore, approximately 4.88 grams of copper(II) sulfide precipitate will form.