What reaction will ensue when sulfur trioxide is added to excess barium hydroxide solution? Question:

If $\ce{SO3}$ was added to an excess solution of $\ce{Ba(OH)2}$, what would be the resulting reaction?

Thoughts:
Initially, I suppose it would form barium hydrogen sulfate, which slowly precipitates barium sulfate in water as the hydroxide is present in excess.
$$\ce{2SO3 + Ba(OH)2 &lt;=&gt; Ba(HSO4)2 \\ Ba(HSO4)2 + Ba(OH)2 &lt;=&gt; 2BaSO4 + 2H2O}$$
Is this correct?
Basically yes.
Sulfur trioxide reacts with water to form sulfuric acid (this is extremely dangerous!  do not try it at home) in a strongly exothermic reaction which, unless done under very precise control, produces a fine mist of concentrated sulfuric acid vapor (not nice).
$$\ce{SO3 + H2O -&gt; H2SO4}$$
If you are doing this with an excess of barium hydroxide then the sulfuric acid will be deprotonated in two steps as you suggest.
$$\ce{H2SO4 + OH- &lt;=&gt; HSO4- + H2O}$$ $$\ce{HSO4- + OH- &lt;=&gt; SO4^2- + H2O}$$
Barium sulfate will precipitate from the solution, driving the reaction to completion.
$$\ce{Ba^2+ (aq) + SO4^2- (aq) -&gt; BaSO4 (s)}$$
I couldn't find any mention of barium hydrogen sulfate as a compound anywhere so I take this to mean that it is not stable and therefore will not precipitate.