\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\\
pthh:2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)
0,4----> 1,2--------->0,4------->0,6
\(V_{H_2}=0,6.22,4=13,44\left(l\right)\\
C\%_{HCl}=\dfrac{1,2.36,5}{200}.100\%=21,9\%\\
m_{\text{dd}.sau.p\text{ư}}=10,8+200-0,6.2=209,6\left(g\right)\\
C\%_{AlCl_3}=\dfrac{133,5.0,4}{209,6}.100\%=25,477\%\)