Write a balanced symbol equation for the reaction between copper carbonate and hydrochloric acid.

To write a balanced symbol equation for the reaction between copper carbonate and hydrochloric acid, we can follow these steps:

Step 1: Identify the Reactants

The reactants in this reaction are:

1. Copper carbonate ($\text{CuCO}_3$)
2. Hydrochloric acid ($\text{HCl}$)

Step 2: Determine the Products

The products formed from the reaction of copper carbonate with hydrochloric acid are:

1. Copper chloride ($\text{CuCl}_2$)
2. Water ($\text{H}_2\text{O}$)
3. Carbon dioxide ($\text{CO}_2$)

Step 3: Write the Unbalanced Equation

The unbalanced equation for the reaction is: $\text{CuCO}_3 + \text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

Step 4: Balance the Equation

To balance the equation:

• On the left, we have 1 copper (Cu), 1 carbonate (CO₃), and 1 hydrogen (from HCl).
• On the right, we need 2 chloride ions (Cl) to balance with copper chloride (CuCl₂), thus 2 HCl are required.

So the balanced equation will be: $\text{CuCO}_3 + 2 \text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} + \text{CO}_2$

Final Answer

The balanced symbol equation for the reaction between copper carbonate and hydrochloric acid is: $\text{CuCO}_3 + 2 \text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} + \text{CO}_2$