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CHEMISTRY, M3 EQ-Bank 28v2

A student stirs 3.50 g of copper (II) nitrate powder into 150.0 mL of 1.50 mol L\(^{-1}\) sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

  1. Write a balanced chemical equation for the reaction.   (1 mark)
  2. Calculate the theoretical mass of precipitate that will be formed.   (4 marks)

    The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
  3. Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error.   (2 marks)
Show Answers Only

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(2.51 \text{ g}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a.    \(\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}\)

b.    \(\ce{n(Cu(NO3)2) = \frac{m}{M} = \frac{3.50}{63.55 + 2 \times (14.01 + 16.00 \times 3)} = \frac{3.50}{187.57} = 0.01866 \text{ mol}}\)

\(\ce{n(NaCl) = c \times V = 1.50 \times 0.150 = 0.225 \text{ mol}}\)

\(\ce{Cu(NO3)2} \text{ is the limiting reagent}\)

\(\ce{n(CuCl2) = n(Cu(NO3)2) = 0.01866 \text{ mol}}\)

\(\ce{m(CuCl2) = n \times M = 0.01866 \times (63.55 + 2 \times 35.45) = 0.01866 \times 134.45 = 2.51 \text{ g}}\)

c.    One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.