Suppose 0.708g of copper (II) acetate is dissolved in 50. mL of a 46.0mM aqueous solution of sodium chromate.

Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the copper (II) acetate is dissolved in it.

The final molarity of acetate anion in the solution is 0.0046 moles

Explanation:

The balanced equation is

Cu (C₂H₃O₂) ₂ + Na₂CrO₄ = CuCrO₄ + 2Na (C₂H₃O₂)

Therefore one mole of Cu (C₂H₃O₂) ₂ react with one mole of Na₂CrO₄ to form one mole of CuCrO₄ and two moles of Na (C₂H₃O₂)

Mass of copper (II) acetate present = 0.708 g

Volume of aqueous sodium present = 50 mL

Molarity of sodium chromate = 46.0 mM

Therefore

Number of moles of sodium chromate present = (50 mL/1000) * 46/1000 = 0.0023 M

Number of moles of copper (II) acetate present = 181.63 g/mol

number of moles of copper (II) acetate present = (0.708 g/181.63 g/mol) = 0.0039 moles

Therefore 0.0039 moles of Cu (C₂H₃O₂) ₂ * (2 moles of Na (C₂H₃O₂)) / 1 Cu (C₂H₃O₂) ₂) = 0.00779 moles of Na (C₂H₃O₂)

also 0.0023 moles of Na₂CrO₄ * (2 moles of Na (C₂H₃O₂)) / 1 Na₂CrO₄) = 0.0046 moles of Na (C₂H₃O₂)

Therefore the Na₂CrO₄ is the limiting reactant and 0.0046 moles of Na (C₂H₃O₂) or acetate anion is formed