Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate: [tex]m_{(AcO)_2Cu} = 0.972 g[/tex]
Volume of the sodium chromate solution: [tex]V_{Na_2CrO_4} = 150.0 mL[/tex]
Molarity of the sodium chromate solution: [tex]c_{Na_2CrO_4} = 0.0400 M[/tex]
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
[tex](CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)[/tex]
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
[tex]n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol[/tex]
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
[tex]n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol[/tex]
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
[tex]Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)[/tex]
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
[tex]n_{(AcO)_2Cu} = 0.0053515 mol[/tex]
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
[tex]n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol[/tex]
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
[tex]c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M[/tex]
