$2Al+6H_2SO_{4(đ)}\buildrel{{t^o}}\over\longrightarrow Al_2(SO_4)_3+3SO_2+6H_2O$
$n_{Al}=\dfrac{5,4}{27}=0,2(mol)$
$n_{SO_2}=\dfrac{3}{2}n_{Al}=\dfrac{3}{2}×0,2=0,3(mol)$
→ $V_{SO_2}=0,3×22,4=6,72(l)$
$n_{Al_2(SO_4)_3}=\dfrac{1}{2}n_{Al}=\dfrac{1}{2}×0,2=0,1(mol)$
→ $m_{Al_2(SO_4)_3}=0,1×342=34,2(g)$