Preparation of Copper(I) Chloride Weight of Copper: 1.004g Volume of Added Nitri
ID: 901264 • Letter: P
Question
Preparation of Copper(I) Chloride
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, 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
The second and third reactions are as follows:
2) 2HNO3(aq) + Na2CO3(s) -> H2O(l) + CO2(g) + 2NaNO3(aq)
3)Cu(NO3)2(aq) + Na2CO3(s) -> CuCO3(s) + 2NaNO3(aq)
Q2. moles of HNO3 left as calculated from first reaction = 0.014
moles of Cu(NO3)2 produced from the first reaction= moles of limiting reagent Cu = 0.0157
Total weight of added Sodium Carbonate: 3.841g
So, moles of sodium carbonate added= 3.841/105.98= 0.0362
2 moles of HNO3 reacts with 1 mole of sodium carbonate,
So, 0.014 moles of HNO3 will react with 0.007 moles of Na2CO3.
Thus, excess moles of Na2CO3 after second reaction= 0.0362-0.007 = 0.029
Now, in the third reaction
Say moles of Cu(NO3)2 used is x and it is the limiting reagent. Then, moles of Na2CO3 used in the third reaction will also be x.
Thus, total moles of Na2CO3 used in second and third reaction are = (0.007+x)
total mass of sodium carbonate needed for the second and third reactions= (0.007+x)*105.9888= (Calculate the value using the value of x )