$m_{HCl}=300.10,95\%=32,85g$
$⇒n_{HCl}=32,85/36,5=0,9mol$
$PTHH :$
$Al_2O_3+6HCl\to 2AlCl_3+3H_2O$
$\text{Theo pt :}$
$n_{Al_2O_3}=1/6.n_{HCl}=1/6.0,9=0,15mol$
$⇒m_{Al_2O_3}=0,15.102=15,3g$
$n_{AlCl_3}=1/3.n_{HCl}=1/3.0,9=0,3mol$
$m_{dd spư}=300+15,3=315,3g$
$⇒C\%_{AlCl_3}=\dfrac{0,3.133,5}{315,3}=12,7\%$