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How did they come up with this completed reaction

 
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Jun20-12, 03:02 PM   #1
 

How did they come up with this completed reaction


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+ ??

Thanks
 
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Jun20-12, 03:59 PM   #2
 
Quote by lechatelier View Post
Here is my attempt, but I don't see how to get it like in the above reaction
Na2CO3 + HCl ⇔ NaCl + HCO3
There is a Na missing.

Quote by lechatelier View Post
The HCO3 breaks down as follows:

HCO3 + H2O ⇔ H3O+ + CO2
Where does the charge comes from?
 
Jun20-12, 04:25 PM   #3
 
Admin
Quote by lechatelier View Post
So my question is how did they only get H2O + CO2 in the top reaction without the hydronium H3O+ ??
Apart from what DrS wrote, solution of HCl contains plenty of H+.
 
Jun21-12, 03:15 AM   #4
 

How did they come up with this completed reaction


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.
 
Jun21-12, 03:34 AM   #5
 
Heavymetal:

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 ??
 
Jun21-12, 03:36 AM   #6
 
Quote by DrStupid View Post
There is a Na missing.



Where does the charge comes from?
Are you serious in asking me where the +ive charge on a hydronium ion came from????
 
Jun21-12, 04:14 AM   #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
 
Jun21-12, 05:05 AM   #8
 
Admin
Quote by HeavyMetal View Post
you get a net equation that looks like:

H2CO3 ⇔ H2O + CO2
That's the main equation here, everything else is just an alternative take on the several equilibria present in the solution.
 
Jun21-12, 03:31 PM   #9
 
Quote by lechatelier View Post
Are you serious in asking me where the +ive charge on a hydronium ion came from????
Yes, I do. The net quantity of electric charge must not change.
 
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