Hướng dẫn:

Gọi ${{n}_{HCl}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=a$ (mol)

${{n}_{{{H}_{2}}}}=\frac{1,344}{22,4}=0,06$ (mol)

$2Al+3{{H}_{2}}S{{O}_{4}}\to A{{l}_{2}}{{(S{{O}_{4}})}_{3}}+3{{H}_{2}}$

            a         $\to $             a

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

a             $\to $     $\frac{1}{2}a$

theo bài cho : $a+\frac{1}{2}a=0,06$ $\Rightarrow a=0,04$ (mol)

                      Chọn đáp án B.