Answer to Question #101415 in General Chemistry for Kalina

Question

CuCl2+Na2CO3->CuCO3+2NaCl

Answer 1

The reaction between Copper (II) Chloride and Sodium Carbonate in an aqueous medium is the combination of the four ions resulting from these compounds. The reaction between the two finally yields Copper Carbonate and Sodium Chloride. It can be represented as:

CuCl₂ + Na₂CO₃ $\to$ CuCO₃ + 2NaCl

A representation of the ions involved is given below:

In the aqueous solution there are ions of Cu²⁺, Na⁺, CO3^2- and Cl^-.

The Cu²⁺ ions combine with the CO₃^2- ions to give CuCO₃.

Cu²⁺ (aq) + CO3^2- (aq) $\to$ CuCO₃(s)

The Na⁺ ions combine with Cl^-1 ions to give NaCl.

Na⁺ + Cl^- $\to$ NaCl