\(n_{Al}=\dfrac{5,67}{27}=0,21\left(mol\right)\)
PTHH: 2Al + 6HCl ---> 2AlCl3 + 3H2
0,21-->0,63----->0,21--->0,315
\(\rightarrow\left\{{}\begin{matrix}V_{H_2}=0,315.22,4=7,056\left(l\right)\\m_{AlCl_3}=0,21.133,5=28,035\left(g\right)\end{matrix}\right.\)