Copper (I) Chloride Reaction Sequence

Question

Copper (I) chloride is prepared from metallic copper, through a series of chemical reactions. In the lab one will make copper(II), remove the excess oxidizing agent (nitrate ion), and then reduce the copper(II) ion to copper(I) by the addition of excess copper metal. The chemical reactions carried out are shown below:

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 + Na2CO3 --> CuCO3 + 2 NaNO3

4) CuCO3 + 2HCL --> CuCl2 + H2O + CO2

5) CuCl2 + Cu --> 2CuCl

Copper metal reacts with dilute (10M) nitric acid by the following reaction:

3Cu(s) + 8HNO3(aq) → 3Cu(NO3)2(aq) + 2NO(g) + 4H₂O(l)

If this reaction took place rather than the first reaction, would the yield of CuCul be affected, assuming you started with the same amount of copper metal? Explain your answer.

Answer 1

The yield of CuCl should not be affected since the mole ratios for Cu(s) and CuCl are the same in both reactions.

In the first reaction 1 mole of Cu(s) produces 1 mole Cu(NO3)2, a 1:1 mole ratio, which ultimately produces 2 mols CuCl

In the alternate reaction 3 moles Cu(s) produces 3 mols Cu(NO3)2, a 1:1 mole ratio, which would then ultimately produce 2 mols CuCl