Paper 1 H - SAMPLE SET 1 Q5
In both reactions one of the products is copper chloride.
5.1) Describe how a sample of copper chloride crystals could be made from copper carbonate and dilute hydrochloric acid. [4 marks]
5.2) A student wanted to make 11.0 g of copper chloride.
The equation for the reaction is:
CuCO3 + 2HCl → CuCl2 + H2O + CO2
Relative atomic masses, Ar: H = 1; C = 12; O = 16; Cl = 35.5; Cu = 63.5
Calculate the mass of copper carbonate the student should react with dilute hydrochloric acid to make 11.0 g of copper chloride.
[4 marks]
Mass of copper carbonate = _______________ g
5.3) The percentage yield of copper chloride was 79.1 %.
Calculate the mass of copper chloride the student actually produced.
[2 marks]
Actual mass of copper chloride produced = _________________ g
5.4) Look at the equations for the two reactions:
Reaction 1 CuCO3(s) + 2HCl(aq) → CuCl2(aq) + H2O(l) + CO2(g)
Reaction 2 CuO(s) + 2HCl(aq) → CuCl2(aq) + H2O(l)
Reactive formula masses: CuO = 79.5; HCl = 36.5; CuCl2 = 134.5; H2O = 18
The percentage atom economy for a reaction is calculated using:
Percentage atom economy = _________________ %
5.5) The atom economy for Reaction 1 is 68.45 %.
Compare the atom economies of the two reactions for making copper chloride.
Give a reason for the difference. [1 mark]
(Total for Question 5 = 14 marks)