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How did they come up with this completed reaction

  1. Jun 20, 2012 #1
    This is what the completed reaction looks like
    Na2CO3 + 2HCl ⇔ 2NaCl + H2O + CO2

    My question is how did they get the products in the above reaction.

    Here is my attempt, but I don't see how to get it like in the above reaction
    Na2CO3 + HCl ⇔ NaCl + HCO3

    The HCO3 breaks down as follows:

    HCO3 + H2O ⇔ H3O+ + CO2

    So my question is how did they only get H2O + CO2 in the top reaction without the hydronium H3O+ ??

  2. jcsd
  3. Jun 20, 2012 #2
    There is a Na missing.

    Where does the charge comes from?
  4. Jun 20, 2012 #3


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    Staff: Mentor

    Apart from what DrS wrote, solution of HCl contains plenty of H+.
  5. Jun 21, 2012 #4
    Well, the first equation you gave us is balanced. The reactants and products both contain:

    3 O
    2 Na
    2 Cl
    2 H
    1 C

    However, your second equation is missing a coefficient in front of one of the reactants. You wrote: Na2CO3 + HCl ⇔ NaCl + HCO3. This is not balanced properly, because on the reactant side you have 2 Na, and on the product side you only account for 1 Na. If you add a two in front of the HCl in the second equation, you get the correct products for the completed reaction of:

    Na2CO3 + 2HCl ⇔ 2NaCl + H2O + CO2.
    Last edited: Jun 21, 2012
  6. Jun 21, 2012 #5

    It's not a question about balancing!!

    I could balance the equation no problem. The question is why the reaction doesn't go this route:

    Na2CO3 + 2HCl ⇔ 2NaCl + H2CO3

    The H2CO3 breaks down as follows:

    H2CO3+ H2O ⇔ H3O+ + HCO3

    The HCO3 breaks down as follows:

    HCO3+ H2O ⇔ H3O+ + CO3

    So I got H3O+ + CO3 but why is it that the correct reaction has H2O + CO2 (at the top, at the beginning of this post) without the hydronium ion H3O+ and CO3 ??
  7. Jun 21, 2012 #6
    Are you serious in asking me where the +ive charge on a hydronium ion came from????
  8. Jun 21, 2012 #7
    I believe it goes:

    H2CO3 + 2H2O ⇔ HCO3- + "H3O+" + H2O
    HCO3- + "H3O+" + H2O ⇔ CO32- + 2"H3O+"
    CO32- + 2"H3O+" ⇔ 3H2O + CO2

    I say "H3O+" because it doesn't actually take shape of the hydronium ion here. So if you cancel out the 2 H2O from the beginning, and the 2 out of the 3 in the end, you get a net equation that looks like:

    H2CO3 ⇔ H2O + CO2

    However, it probably looks more like this:

    H2CO3 ⇔ HCO3- + H+(aq)
    HCO3- + H+(aq) ⇔ CO32- + 2H+(aq)
    CO32- + 2H+(aq) ⇔ H2O + CO2
  9. Jun 21, 2012 #8


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    Staff: Mentor

    That's the main equation here, everything else is just an alternative take on the several equilibria present in the solution.
  10. Jun 21, 2012 #9
    Yes, I do. The net quantity of electric charge must not change.
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