a/
$n_{Al}=5,4/27=0,2mol$
$2Al+6HCl\to 2AlCl_3+3H_2$
Theo pt:
$n_{HCl}=3n_{Al}=0,6mol$
$⇒m_{HCl}=0,6.36,5=21,9g$
b/
$m_{ddHCl}=\dfrac{21,9}{14\%}=156,43g$
Theo pt:
$n_{AlCl_3}=n_{Al}=0,2mol$
$n_{H_2}=3/2.n_{Al}=0,3mol$
$m_{AlCl_3}=0,2.133,5=26,7g$
$m_{ddsau}=m_{ddHCl}+m_{Al}-m_{H_2}$
$=156,43+5,4-0,3.2=161,23g$
$C\%muối=\dfrac{26,7}{161,23}.100\%=16,56\%$