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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.

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Answer:

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

#b. 0.0231 moles COâ‚‚

Explanation:

  1. Carbonates reacts with acid to produce carbon dioxide and water as products.
  • The reaction between sodium carbonate and hydrochloric acid yields sodium chloride, carbon dioxide and water.

The balanced equation for the reaction is;

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

Part B

  • We are required to calculate the number of moles of COâ‚‚ if 23.1 ml of Naâ‚‚CO₃ were used.

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

  • Volume of Naâ‚‚CO₃ is 23.1 mL or 0.0231 moles
  • Molarity of concentration of Naâ‚‚CO₃ is 1.0 M

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 COâ‚‚ produced.

  • From the equation 1 mole of Naâ‚‚CO₃ reacts with the acid to produce 1 mole of COâ‚‚.

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

Therefore; Moles of COâ‚‚ produced is 0,0231 moles

   = 0.0231 moles CO₂

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