$n_{Al}=3,6/27=\dfrac{2}{15}mol$
$a/2Al+6HCl\to 2AlCl_3+3H_2↑$
$\text{b/Theo pt :}$
$n_{AlCl_3}=n_{Al}=\dfrac{2}{15}mol$
$⇒m_{AlCl_3}=\dfrac{2}{15}.133,5=17,8g$
$\text{c/Theo pt :}$
$n_{H_2}=3/2.n_{Al}=3/2.\dfrac{2}{15}=0,2mol$
$⇒V_{H_2}=0,2.22,4=4,48l$