Question

ionic equation between coper chloride and sodioum carbonate

Answers

GPT-4o mini
To write the ionic equation for the reaction between copper(II) chloride (CuCl₂) and sodium carbonate (Na₂CO₃), we first need to identify the products of the reaction. When these two compounds react, they undergo a double displacement reaction, resulting in the formation of copper(II) carbonate (CuCO₃) and sodium chloride (NaCl).

The balanced molecular equation is:

\[
\text{CuCl}_2 (aq) + \text{Na}_2\text{CO}_3 (aq) \rightarrow \text{CuCO}_3 (s) + 2 \text{NaCl} (aq)
\]

Next, we can break all the soluble ionic compounds into their constituent ions to write the full ionic equation. The sodium chloride (NaCl) dissociates in solution, while copper(II) carbonate (CuCO₃) is insoluble and will remain as a solid.

The full ionic equation is:

\[
\text{Cu}^{2+} (aq) + 2 \text{Cl}^- (aq) + 2 \text{Na}^+ (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s) + 2 \text{Na}^+ (aq) + 2 \text{Cl}^- (aq)
\]

In order to obtain the net ionic equation, we can cancel the spectator ions (ions that appear on both sides of the equation). In this case, the sodium ions (\( \text{Na}^+ \)) and chloride ions (\( \text{Cl}^- \)) are spectator ions.

Thus, the net ionic equation is:

\[
\text{Cu}^{2+} (aq) + \text{CO}_3^{2-} (aq) \rightarrow \text{CuCO}_3 (s)
\]

This equation represents the formation of solid copper(II) carbonate from its ions in solution.