Preparation of Copper(I) Chloride Reactions: 1) Cu(S) + 4HNO 3 (aq) -> Cu(NO 3 )
ID: 901376 • Letter: P
Question
Preparation of Copper(I) Chloride
Reactions:
1) Cu(S) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
2) 2HNO3(aq) + Na2CO3(s) -> H2O(l) + CO2(g) + 2NaNO3(aq)
3)Cu(NO3)2(aq) + Na2CO3(s) -> CuCO3(s) + 2NANO3(aq)
4)CuCO3(s) + 2HCL(aq) -> CuCl2(aq) + H2O(l) + CO2(g)
5)CuCl2(aq) + Cu(s) -> 2CuCl(s)
Report:
Weight of Copper: 1.004g
Volume of Added Nitric Acid: 4.9mL
Total weight of added Sodium Carbonate: 3.841g
Weight of watch glass and filter paper: 51.740g
Weight of watch glass, filter paper, and CuCl Precipitate: 53.513g
Experimental Yield of CuCl: 1.775g
Theoretical Yield of CuCl: 3.55g
Percent Yield of CuCl: 50%
1) Based on the amounts of copper metal and nitric acid used in the first reaction, calculate the number of moles of HNO3 there are in excess. Concentrated nitric acid has a concentration of 15.8 M.
2) Using the moles of HNO3 you calculated and the moles of Cu(NO3)2 produced from the first reaction, calculate the total mass of sodium carbonate needed for the second and third reactions. Did you add enough sodium carbonate in the experiment?
3) What observation suggests that copper was added in excess during the last reaction?
4) Why might a percent yield of less than 100% for this experiment be obtained?
5) Copper metal reacts with dilute nitric acid by the following reaction.
3Cu(s) +8HNO3(aq) + 2NO(g) + 4H20(l)
If this reaction took place rather than the first reaction, would your yield of CuCL be affected assuming you started with the same amount of copper metal? Explain your answer.
Explanation / Answer
1) Cu(S) + 4HNO3(aq) -> Cu(NO3)2(aq) + 2NO2(g) + 2H2O(l)
millimoles of Cu =- 1.004 g/63.5= 15.8 millimoles
millimoles of HNO3 = 4.9 x 15.8 M = 77.42 millimoles
excess moles of HNO3 = 77.42 - 15.8x4 = 14.22 millimoles are unreacted.
millimoles of Co(NO3) Formed = 15.84 millimoles
2) 2HNO3(aq) + Na2CO3(s) -> H2O(l) + CO2(g) + 2NaNO3(aq)
The millimoles of Na2CO3 needed = 1/2 ( millimoles of unused HNO3 ) =1/2 x14.22= 7.11 millimoles
Millimoles of Na2CO3 added = 3.841/106 x1000 = 36.2 miilimoles
so we have enough sodium carbonate.
3)Cu(NO3)2(aq) + Na2CO3(s) -> CuCO3(s) + 2NANO3(aq)
4)CuCO3(s) + 2HCL(aq) -> CuCl2(aq) + H2O(l) + CO2(g)
5)CuCl2(aq) + Cu(s) -> 2CuCl(s)