${{n}_{Al}}=\frac{{{m}_{Al}}}{{{M}_{Al}}}=\frac{4,05}{27}=0,15\left( mol \right)$

$2Al+6HCl\to 2AlC{{l}_{3}}+3{{H}_{2}}$

 0,15             $\to $            0,225       (mol)

Theo phương trình ta có:

 ${{n}_{{{H}_{2}}}}=\frac{3.{{n}_{Al}}}{2}=\frac{3.0,15}{2}=0,225\left( mol \right)$

$\Rightarrow {{V}_{{{H}_{2}}}}=22,4.{{n}_{{{H}_{2}}}}=22,4.0,225=5,04\left( l \right)$

$\Rightarrow $Chọn đáp án A.