${{n}_{Mg}}=\frac{{{m}_{Mg}}}{{{M}_{Mg}}}=\frac{6}{24}=0,25\left( mol \right)$

$Mg+2HCl\to MgC{{l}_{2}}+{{H}_{2}}$

0,25             $\to $              0,25             (mol)

Theo phương trình ta có: ${{n}_{{{H}_{2}}lt}}={{n}_{Mg}}=0,25\left( mol \right)$

$\Rightarrow {{n}_{{{H}_{2}}pu}}=0,9.{{n}_{{{H}_{2}}lt}}=0,25.0,9=0,225\left( mol \right)$

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

$\Rightarrow $ Chọn đáp án D