When $\mathrm{AlCl}_3$ reacts with $\mathrm{NaOH}$, it forms sodium meta aluminate $\left(\mathrm{NaAlO}_2\right)$. This reaction does not give gaseous product.
$\mathrm{AlCl}_3+4 \mathrm{NaOH} \longrightarrow \mathrm{NaAlO}_2$ (Sod. meta aluminate (soluble)) $+2 \mathrm{H}_2 \mathrm{O}+3 \mathrm{NaCl}$