$a/$
$2Al +6HCl→2AlCl3+3H2$
$b/$
$n_{Al} =5,4/27=0,2mol$
$n_{H_2}=3/2.n_{Al}=0,3mol$
$V_{H_2}=0,3.22,4=6,72l$
$c/$
$n_{AlCl_3}=n_{Al} =0,2mol$
$m_{AlCl_3}=0,2.133,5=26,7g$
$d/$
$n_{HCl}=3.n_{Al}=3.0,2 =0,6mol$
$m_{HCl}=0,6.36,5=21,9g$
$⇒C\%_{HCl}=\dfrac{21,9}{200}.100\%=10,95\%$