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When gases leave the catalytic reactor in the contact process of sulfuric acid generation, both $\ce{SO2}$ and $\ce{SO3}$ will be present. According to descriptions I have read, the trioxide reacts with the water spray, but the dioxide does not, and instead rises as a gas and is recirculated into the system.

I thought that sulfur dioxide was highly soluble in water. If so, should it not dissolve into the water spray and thereby contaminate the oleum?

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You are true that $\ce{SO2}$ is very soluble in water but it depends on the temperature. Also, $\ce{SO3}$ reacts with water then dissolves in it in a very exothermic way. My understanding is that, if the quantity of water is well adjusted, you will actually get hot concentrated sulfuric acid and sulfur dioxide is not quite soluble in it (ca. 8 g/kg at 120 °C). So $\ce{SO3}$ reacts with water and sulfur dioxide gas can be recirculated.

HTH

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$\ce{SO3}$ reacts rapidly and exothermically with $\ce{H2O}$, forming the liquid $\ce{H2SO4}$, sulfuric acid. This removes the $\ce{SO3}$ from the mixed gases, allowing yet more to react.

In fact, the resulting acid can contain more $\ce{SO3}$ than $\ce{H2O}$. This is called oleum or fuming sulfuric acid.

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    $\begingroup$ This doesn't really answer the question. The question is about why $\ce{SO2}$ does not dissolve in the oleum. $\endgroup$ – bon Aug 8 '16 at 8:02

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