inorganic chemistry - HCl synthesis from sulfuric acid and sodium salt

I'd like know if this is right. The reaction between $\ce{NaCl}$ and $\ce{H_2SO_4}$ should yield $\ce{HCl}$ and sodium sulfate. This is a double substitution, true? No oxidation took place here. The thing is, when I tried the reaction in the lab, the resulting solution (gas bubbled into water) seemed to be sulfuric acid. Simply enough, the solution didn't react with aluminum. I presume that $\ce{SO_2}$ was formed from the oxidation the sulfur (it a little bit like it).

What am I doing wrong? Shouldn't the substitution take place instead?

Process:

$ \ce{NaCl + H_2SO_4 \rightarrow NaHSO_4 + HCl}\\ \ce{NaCl + NaHSO_4 \rightarrow Na_2SO_4 + HCl} $

Adding:

$ \ce{2\space NaCl + H_2SO_4 \rightarrow Na_2SO_4 + 2\space HCl } $

Is this right?

Answer

Ideally, that's the reaction. Realistically, there are a few speed bumps.

First, it occurs stepwise, with the acid salt sodium bisulfate as an intermediate. The first stage happens at room temp between equal proportions of salt and sulfuric acid. The second stage requires temperatures of 200*C (in turn requiring water-free concentrations of acid and dry salt) and an additional equivalent of salt that reacts with the bisulfate.

You did say you were using excess sulfuric acid; if true, you'll react all the salt but what'll be left in the flask is sodium bisulfate, not sodium sulfate. Since you're interested in the HCl gas produced and not the sulfate salt, this isn't a big deal, but I just wanted you to be clear that one mole of sulfuric acid to two moles of salt isn't going to get you the expected 2 moles of HCl unless you're heating a "water-free" combination of the reactants.

Second, if you have too much water in the reaction, the HCl won't bubble out; it'll happily sit in solution. So, dry salt and 10M or higher concentration of sulfuric acid are recommended (as in Nicolau's YouTube video, 18M is plenty strong enough to do the job, and they make and sell it up to 98% concentrations). If you're stuck with 1M, you can heat the flask to boiling, which evaporates some water and increases the acid concentration, liberating the HCl gas (sulfuric acid won't boil below 300*C, so you're fine distilling the HCl like this until you're out of water and the thermometer starts to climb).

Following these steps and bubbling the gas produced through water, what you produce should definitely be hydrochloric acid. Obviously the molar strength of it (and thus its ability to react with test materials as was also indicated by Nicolau) will depend on the relative proportions and amount of reactants, the volume of your trap water, and how much gas is actually absorbed by the water.