When 1.0 m aqueous solutions of sodium carbonate and hydrochloric acid react it produces a carbon dioxide, water and sodium chloride (2 points) a. write the balanced chemical equation for the reaction described above (include states of matter) b. calculate the number of moles of carbon dioxide formed if 23.1 ml of sodium carbonate reacts with excess hydrochloric acid.

#a.  Na₂CO₃(aq) + 2HCl (aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

#b. 0.0231 moles CO₂

Explanation:

The balanced equation for the reaction is;

Na₂CO₃(aq) + 2HCl (aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)

Part B

First, we need to calculate the number of moles of Na₂CO₃

But;

Number of moles = Volume × Concentration

Therefore;

Moles of Na₂CO₃ = 0.0231 L× 1.0

                             = 0.0231 moles Na₂CO₃

Second, we can determine the moles of CO₂ produced.

This means, the mole ratio of Na₂CO₃ : CO₂ is 1 : 1

Therefore; Moles of CO₂ produced is 0,0231 moles

   = 0.0231 moles CO₂

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