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Chemistry

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Step-by-step solution

3 steps
  1. Identify the reactants and products.

    The reactants are sodium carbonate (Na2CO3\text{Na}_2\text{CO}_3) and hydrochloric acid (HCl\text{HCl}).
    The products of the reaction between a carbonate and an acid are a salt, water, and carbon dioxide. In this case, the salt formed is sodium chloride (NaCl\text{NaCl}).

  2. Write the unbalanced chemical equation.

    Na2CO3(aq)+HCl(aq)→NaCl(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow \text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})
  3. Balance the chemical equation.

    • Balance sodium (Na): There are 2 Na atoms on the left, so place a coefficient of 2 in front of NaCl\text{NaCl} on the right.
    Na2CO3(aq)+HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3(\text{aq}) + \text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})
    • Balance chlorine (Cl): Now there are 2 Cl atoms on the right, so place a coefficient of 2 in front of HCl\text{HCl} on the left.
    Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})
    • Check hydrogen (H), carbon (C), and oxygen (O):
      • H: 2 atoms on the left (from 2HCl2\text{HCl}) and 2 atoms on the right (from H2O\text{H}_2\text{O}). Balanced.
      • C: 1 atom on the left (from Na2CO3\text{Na}_2\text{CO}_3) and 1 atom on the right (from CO2\text{CO}_2). Balanced.
      • O: 3 atoms on the left (from Na2CO3\text{Na}_2\text{CO}_3) and 1 atom (from H2O\text{H}_2\text{O}) + 2 atoms (from CO2\text{CO}_2) = 3 atoms on the right. Balanced.

    The balanced chemical equation is:

    Na2CO3(aq)+2HCl(aq)→2NaCl(aq)+H2O(l)+CO2(g)\boxed{\text{Na}_2\text{CO}_3(\text{aq}) + 2\text{HCl}(\text{aq}) \rightarrow 2\text{NaCl}(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})}
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