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

Question

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Solution

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 (CuCO3 \text{CuCO}_3 )
  2. Hydrochloric acid (HCl \text{HCl} )

Step 2: Determine the Products

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

  1. Copper chloride (CuCl2 \text{CuCl}_2 )
  2. Water (H2O \text{H}_2\text{O} )
  3. Carbon dioxide (CO2 \text{CO}_2 )

Step 3: Write the Unbalanced Equation

The unbalanced equation for the reaction is: CuCO3+HClCuCl2+H2O+CO2 \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: CuCO3+2HClCuCl2+H2O+CO2 \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: CuCO3+2HClCuCl2+H2O+CO2 \text{CuCO}_3 + 2 \text{HCl} \rightarrow \text{CuCl}_2 + \text{H}_2\text{O} + \text{CO}_2

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