Answer:
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
