$n_{Al_2(SO_4)_3}=\dfrac{51,3}{342}=0,15mol$
$a.PTHH :$
$2Al+3H_2SO_4\to Al_2(SO_4)_3+3H_2↑$
$\text{b.Theo pt :}$
$n_{Al}=2.n_{Al_2(SO_4)_3}=2.0,15=0,3mol$
$⇒m_{Al}=0,3.27=8,1g$
$\%m_{Al}=100\%-19\%=81\%$
$⇒a=8,1:81\%=10g$
$n_{H_2SO_4}=n_{H_2}=3.n_{Al_2(SO_4)_3}=3.0,15=0,45mol$
$⇒b=m_{H_2SO_4}=0,45.98=44,1g$
$V=V_{H_2}=0,45.22,4=10,08l$