When a NaSO3 solution is added to a CuCl2 solution, a white precipitate is formed. This precipitate is copper(I) sulfite, Cu2SO3.

Chemical Reaction:

The chemical reaction that takes place is:

$$2CuCl_2_{(aq)} + 3Na_2SO_{3(aq)} -> Cu_2SO_{3(s)} + 6NaCl_{(aq)}$$

In this reaction, the copper(II) ions in the CuCl2 solution are reduced to copper(I) ions by the sulfite ions from the NaSO3 solution. The copper(I) ions then react with the sulfite ions to form the white precipitate of copper(I) sulfite.

Additional Information:

The formation of copper(I) sulfite precipitate can be used as a qualitative test for the presence of copper(II) ions in a solution.