$2Al+3H_2SO_4→Al_2(SO_4)_3+3H_2$
$n_{Al}=\dfrac{4,05}{27}=0,15(mol)$
$n_{H_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}×0,15=0,225(mol)$
→ $V_{H_2}=0,225×22,4=5,04(l)$
$n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}×0,15=0,075(mol)$
→ $m_{Al_2(SO_4)_3}=0,075×342=25,65(g)$
(giá trị $x$ là gì thế bạn?)