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In lab, we performed the Solvay process where we had to produce sodium carbonate as the final product. There is a certain step where we put something in the oven and after this, we expect that the product is sodium bicarbonate ($\ce{NaHCO3}$). This product does not get named in the instructions; it is simply referred to as "dry product". After this step, we were asked to add 2 drops of $\ce{HCl}$ to a small amount of the dry product in a watch glass. Finally, we should repeat the watch glass step using $\ce{NaHCO3}$ and $\ce{Na2CO3}$.

After doing these 3 experiments, I observed the same thing (bubbles). I think that the reactions were:

$$\ce{NaHCO3 + HCl -> NaCl + CO2 + H2O}$$

and for the third one (since I added $\ce{HCl}$ drop-wise to $\ce{Na2CO3}$, $\ce{Na2CO3}$ was in excess):

$$\ce{Na2CO3 + HCl -> NaHCO3 + NaCl}$$

and then:

$$\ce{Na2CO3 + HCl -> NaCl + CO2 + H2O}$$

I believe the purpose of the 3 experiments was to identify that the dry product was $\ce{NaHCO3}$, but after adding $\ce{HCl}$, I observed exactly the same thing (bubbles) for both known reagents. So I'm confused as to what should be my answer.

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  • $\begingroup$ Or the reactions should be: Na2CO3(s) -> 2Na+(aq) + CO3-2(aq) , (2) CO3-2(aq) + H+ <--> HCO3-(aq) , (3) HCO3-(aq) + H+ <--> H2CO3(l) ---> H2O(l) + CO2(g) ? $\endgroup$ – chemistrylove Mar 23 '17 at 17:03
  • $\begingroup$ Your 3rd equation is not balanced. You will need 2 moles of hydrochloric acid to get the same products and that's how you can separate sodium carbonate and sodium bicarbonate. NaHCO3 needs one mole of HCl whereas Na2CO3 needs two moles. Also the amount of NaCl produced in Na2CO3 is more than that of NaHCO3. $\endgroup$ – Nilay Ghosh Mar 27 '17 at 3:39
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It seems to me that the purpose of the 3 experiments with the HCl was not to identify the dry product. You can just write the reactions you observed, that was probably the point.

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