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In the reaction $\ce{Na(s) + H2O}$ the products are listed as $\ce{NaOH(aq) + H2 (g)}$, but why would $\ce{Na2O}$ not be a possible product?
I searched and found that the reaction $\ce{Na2O + 2H2O -> 2NaOH}$ occurs as well. Does it have to do with the stability of oxides?
In the usual case every student learns, the reaction of sodium with water is the following — I made note to include the states of matter here as they matter (pun intended).
$$\ce{2 Na (s) + 2 H2O (l) -> 2 Na+ (aq) + 2 OH- (aq) + H2 ^ (g)}\tag{1}$$
In the typical setup, you have lots of water and little sodium. Therefore, whatever you get will be a solution or a precipitated salt. Alkali metals, however, often form very soluble salts and indeed all alkali hydroxides are well soluble. Thus, we will never get to see any solid by-products; all will happen in the solution phase.
Technically, it is possible for an oxide anion to be formed. However, this is when Brønsted and Lowry’s acid-base theory kicks in. Remember that water is amphoteric and can react as a base to form $\ce{H3O+}$ or as an acid to form $\ce{OH-}$. Well, equation $(2)$ expands that scheme.
$$\ce{H3O+ <=>[$K_\mathrm{a,1}$][H+] H2O <=>[$K_\mathrm{a,2}$][H+] OH- <=>[$K_\mathrm{a,3}$][H+] O^2-}\tag{2}$$
The oxide anion is connected to hydroxide and water by a simple acid-base reaction. From the definition of $\mathrm{p}K_\mathrm{a}$ values and the ionic product of water, we know that
$$\mathrm{p}K_\mathrm{a} (\text{acid}) + \mathrm{p}K_\mathrm{b} (\text{conj. base}) = 14\tag{3}$$
Since $K_\mathrm{w}$ doubles as the acidity constant of water, $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = 14$. Likewise it can be shown, that $\mathrm{p}K_\mathrm{a} (\ce{H3O+}) = 0$. This is a difference of $14$ logarithmic units between two subsequent deprotonation steps. The acidity constants of sulfuric acid or phosphoric acid show that while this difference is large, it is not unusual for the conjugate base (if it is acidic) to have an acidity constand orders of magnitude lower than the original acid).
Condensing that into a conclusion, the acidity constant of hydroxide must be even lower still, to the point where we can say practically no oxide anions can be formed in equilibrium. Likewise, considering that there are many more water molecules than hydroxide anions, any oxide accidentally generated will get protonated immediately.
$$\ce{O^2- (s) + H2O (l) -> 2 OH- (aq)}\tag{4}$$
Therefore, it is not feasible for oxide anions to exist in water. This is known as the nivelling effect: no acid more acidic than $\ce{H3O+}$ can survive in aquaeous solution for extended time and no base more basic than $\ce{OH-}$.
Since the entire reaction produces products in solution, sodium oxide is not a possibility.
What compounds are formed in a system that contains sodium, hydrogen and oxygen depends on the pressure, temperature, and the number of atoms of the respective type.
If sodium reacts with an excess of water at standard conditions, dissolved sodium hydroxide and hydrogen are formed.
$$\ce{2 Na + 2 H2O-> 2 Na+_{(aq)} + 2 OH-_{(aq)} + H2}$$
Sodium oxide can be synthesized by reaction of sodium hydroxide with sodium:
$$\ce{2 NaOH + 2 Na -> 2 Na2O + H2}$$
Additionally sodium oxide can react with hydrogen to form sodium hydroxide and sodium hydride under certain conditions. The reaction is reversible:
$$\ce{Na2O + H2 <=> NaOH + NaH}$$
From the first two reaction equations above we can conclude that we also could obtain sodium oxide by reaction of sodium with water.
Sodium oxide dissolves in water to form sodium hydroxide:
$$\ce{Na2O(aq) + H2O(l) -> 2NaOH(aq)}$$
So even if sodium oxide is formed, it would be immediately converted to sodium hydroxide.
