Products of decarboxylation of sodium carboxylate
The first reaction is:
$$\ce{CH3COONa + NaOH -&gt; Na2CO3 + CH4}$$
According to Encyclopedia of Inorganic Chemistry vol. 2:

The obtained methane in this process is not pure and may contain as
much as 8 % hydrogen and 10% other hydrocarbons such as the chloride
(to remove  acetylene) and concentrated sulfuric acid (to remove
ethylene and moisture). The purified  methane still contains hydrogen,
which may be removed as water by boiling at 110 °C the  excess oxygen
is absorbed concentrated sulfuric acid.

The second reaction doesn't make much sense to me with $\ce{BaO}$. If this reaction is happening in an aqueous medium, then $\ce{BaO}$ would probably have reacted to form $\ce{Ba(OH)2}$*. So I think you could justify representing the second reaction as:
$$\ce{CH3COONa + Ba(OH)2 -&gt; CH4 + NaCO3 + BaCO3}$$
This is also how the aforementioned book represents the reaction, stating:

Heating sodium acetate with barium hydroxide, instead of sodium hydroxide forms nearly pure methane:
Sodium Acetate + Barium Hydroxide $\rightarrow$ Methane + Sodium Carbonate + Barium Carbonate
$$\ce{CH3COONa + Ba(OH)2 -&gt; 2CH4 + NaCO3 + BaCO3}$$

So I guess the increased purity, in this case, is just due to the different amount of other products produced during the reaction.