Hướng dẫn:

${{n}_{Al}}=\frac{{{m}_{Al}}}{{{M}_{Al}}}=\frac{3,78}{27}=0,14$(mol)

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

0,14$\to $ 0,21 (mol)

Theo phương trình :${{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{3}{2}{{n}_{Al}}=\frac{3}{2}.0,14=0,21$ (mol)

${{V}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}}=\frac{0,21}{0,35}=0,6$  lít

                      Chọn đáp án A.