Reaction of barium peroxide forming barium carbonate
Before the precipitation of $\ce{BaCO3}$ starts, the following equilibrium is arranged: 

$$\ce{BaO2 + 2H-O-H &lt;=&gt; H2O2 + Ba^{2+} + 2OH-}$$

$\ce{BaO2}$ is but very slightly soluble in $\ce{H2O}$ because the then occurring hydrolysis produces $\ce{OH^-}$, which raises the pH, and hence stops hydrolysis (and dissolving) of further $\ce{BaO2}$. Simultaneously, by the formed $\ce{OH^-}$, the back reaction is fostered. However, these two blockades may be overcome by adding dilute acid, since this latter neutralizes at once the alkaline product $\ce{Ba(OH)2}(aq)$, and so the equilibrium may be continually shifted to the right side.

Hence, it doesn't matter whether the acid is more or less diluted (nor whether it is strong or weak), provided that acid is continually supplied to the extent that it is consumed during the dissolution/hydrolysis process of $\ce{BaO2}$. An aqueous solution of $\ce{Ba(OH)2}$ is an efficient gas detection or gas scrubbing method for $\ce{CO2}$; hence, by bubbling  $\ce{CO2}$-gas through it for a long time, the $\ce{Ba(OH)2}(aq)$ will easily be transformed into $\ce{BaCO3}(s)$. Of course, a gas frit will be used to spread the gas, and the reaction mixture should be constantly stirred.  

In the textbook of Sarda/Handa/Arora$^1$ this process with $\ce{CO2}$ is shortly mentioned (without any explanations except that ice-cold water should be used, and that $\ce{BaCO3}$ may be removed by filtration, since the desired product is, of course, $\ce{H2O2}$).



$^1$ Sanda/Harda/Arora, Chemistry, part-2, New Dehli, 2016, p. 885.