Fig. 1 shows stages used by a school technician to prepare copper sulfate crystals for an electroplating demonstration. Copper oxide is warmed with dilute s...
Question 1 Report
Fig. 1 shows stages used by a school technician to prepare copper sulfate crystals for an electroplating demonstration. Copper oxide is warmed with dilute sulfuric acid in stage A. Extra copper oxide is added until black solid remains in stage B. The mixture is filtered in stage C and the blue filtrate is heated carefully in stage D. The technician cools the concentrated solution before collecting crystals. At 80°C, 100 g of water dissolves 25 g of copper sulfate. At 20°C, the same mass of water dissolves 8 g of copper sulfate.
(a) Name the salt made in this preparation. [1]
(b) Give two reasons why copper oxide is added until some solid remains. [2]
(c) State why the mixture is filtered at stage C. [1]
(d) Complete the equation: CuO(s) + H2SO4(aq) → .......... + .......... [2]
(e) Describe three steps after stage D that produce dry copper sulfate crystals. [3]
(f) Calculate the minimum volume of 1.00 mol dm-3 sulfuric acid needed to react completely with 4.00 g of copper oxide. Relative formula mass: CuO = 79.5. [3]
(g) Calculate the mass of copper sulfate crystals that forms when 100 g of saturated solution at 80°C is cooled to 20°C. [3]
Answer Details
(a) The salt made is copper sulfate. [1]
(b) Copper oxide is added until some remains to ensure all sulfuric acid has reacted and to avoid excess acid in the copper sulfate solution. Since copper oxide is insoluble, excess solid can be removed by filtration. Any two points gain credit. [2]
(c) Stage C filters off excess unreacted copper oxide solid. [1]
(d) \[\mathrm{CuO(s)+H_2SO_4(aq)\rightarrow CuSO_4(aq)+H_2O(l)}\] [2]
(e) After stage D, allow the concentrated solution to cool so crystals form. Filter the crystals from the remaining solution and dry them between filter papers. Heating should only concentrate the solution, not heat it to dryness. [3]
(f) \[n(\mathrm{CuO})=\frac{4.00\text{ g}}{79.5\text{ g mol}^{-1}}=0.0503\text{ mol}\]
The equation has a \(1:1\) ratio, so \(0.0503\text{ mol}\) of \(\mathrm{H_2SO_4}\) is needed.
\[V=\frac{n}{c}=\frac{0.0503}{1.00}=0.0503\text{ dm}^3=50.3\text{ cm}^3\]
The minimum volume is \(50.3\text{ cm}^3\). [3]
(g) In 100 g water, 25 g dissolves at \(80^\circ\mathrm{C}\), but only 8 g stays dissolved at \(20^\circ\mathrm{C}\).
\[25\text{ g}-8\text{ g}=17\text{ g}\]
The mass of crystals formed is 17 g. [3]
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