Question

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

Answer 1

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