Solution:

$\mathrm{KAl}\left(\underline{\mathrm{S}} \mathrm{O}_{4}\right)_{2} .12 \mathrm{H}_{2} \mathrm{O}$

Let expect oxidation number of S is x.

\[\begin{array}{*{35}{l}}

Oxidation\text{ }number\text{ }of\text{ }K\text{ }=\text{ }+1  \\

~  \\

Oxidation\text{ }number\text{ }of\text{ }Al\text{ }=\text{ }+3  \\

~  \\

Oxidation\text{ }number\text{ }of\text{ }O\text{ }=\text{ }-\text{ }2  \\

~  \\

Oxidation\text{ }number\text{ }of\text{ }H\text{ }=\text{ }+1  \\

\end{array}\]

Then, at that point, we have:

\[\begin{array}{*{35}{l}}

1\left( +1 \right)\text{ }+\text{ }1\text{ }\left( +3 \right)\text{ }+\text{ }2\left( x \right)\text{ }+\text{ }8\left( -\text{ }2 \right)\text{ }+\text{ }24\left( +1 \right)\text{ }+\text{ }12\text{ }\left( -\text{ }2 \right)\text{ }=\text{ }0  \\

~  \\

=\text{ }1\text{ }+\text{ }3\text{ }+\text{ }2x\text{ }-\text{ }16\text{ }+24\text{ }-\text{ }24\text{ }=\text{ }0  \\

~  \\

=\text{ }2x\text{ }-\text{ }12\text{ }=\text{ }0  \\

~  \\

=\text{ }2x\text{ }=\text{ }+12  \\

~  \\

=\text{ }x\text{ }=\text{ }+6  \\

\end{array}\]

Thus, Oxidation number of S is +6.