Answer:
\[\frac{\begin{align}
& C{{l}_{2}}+{{H}_{2}}O\to 2HCl+[O] \\
&
S{{O}_{ & 2}}+[O]\to S{{O}_{3}} \\
\end{align}}{\frac{C{{l}_{2}}+S{{O}_{2}}+{{H}_{2}}O\to
2HCl+S{{O}_{3}}}{{}}}\]
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