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When sodium oxide is added into water, it reacts with water to form $\ce{NaOH}$. $$\ce{Na2O(s) + H2O(l) -> 2 NaOH(aq)}$$

This means that if we were to add an acid, it would be neutralized by the sodium hydroxide. $$\ce{Na2O(s) + H2O(l) + 2 HCl(aq) -> 2 NaCl(aq) + 2 H2O}$$

Is the following equation a totally different reaction pathway, or is it simplified from the reaction pathways above? $$\ce{Na2O(s) + 2 HCl(aq) -> 2 NaCl(aq) + H2O(l)}$$

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    $\begingroup$ Most likely, nobody actually knows the correct path. The question, does the oxide react with water or does it react with protons from HCl. Both must be extremely fast reactions. I think someone has to use to isotope labelling in Na2O and H2O and see what it is isotope of oxygen in NaOH. $\endgroup$
    – M. Farooq
    Apr 13 '20 at 4:57
  • $\begingroup$ It is surprising that there is no related paper on this topic. $\endgroup$
    – M. Farooq
    Apr 13 '20 at 5:10
  • $\begingroup$ First you have to decide what experiment you want to do. Add HCl to solid sodium oxide, or dissolve the sodium oxide in water first? $\endgroup$
    – Karl
    Apr 13 '20 at 6:32
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    $\begingroup$ As water will present at about 55 molar, even in the presence of acid, the first reaction would seem to be be the most likely assuming that it does not exist as Na$_2$O in solution. $\endgroup$
    – porphyrin
    Apr 13 '20 at 7:19
  • $\begingroup$ And then perhaps give a thought to the concentration of your HCl solution. $\endgroup$
    – Karl
    Apr 13 '20 at 7:20
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Both fast reactions happen in parallel, acting as "first comes, first served".

Concentration of water molecules ( $\pu{55 mol/L}$ in water) is much higher than $\ce{[H3O+]}$ for diluted acids, so oxide anions meet water molecules much more frequently.

If the oxide anion meets water molecule, it will not wait for hydronium. Additionally, hydronium ions would be rather depleted by reaction with hydroxide ions next to the oxide surface.

Be aware that $\ce{Na+}$ ions are just bystander/spectator ions, so it is more illustrative to write equations as ionic equations:

The main reaction: $$\ce{O^2− + H2O -> 2 OH−}\tag{1}$$

The secondary alternative reaction: $$\ce{O^2− + H3O+ -> OH− + H2O}\tag{2}$$

The final reaction: $$\ce{H3O+ + OH− -> 2 H2O}\tag{3}$$

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Both are equally valid (different) pathways though indistinguishable in real life.

Sodium oxide is essentially an anhydrous version of sodium hydroxide. Your second pathway is valid as sodium oxide being a solid base would lose the oxide ion when in contact with an acid like HCl.

Your first pathway is equally valid, where the sodium oxide dissolves in water first then reacts with HCl.

However, these two both happen so fast that they are basically indistinguishable in normal settings. The only situation where I perceive only one happening would be in high temperatures where HCl is considered molten and not aqueous, then the oxide does not dissolve into water but reacts straightaway.

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