Hướng dẫn:

${{m}_{c{{t}_{{{H}_{2}}S{{O}_{4}}}}}}=\frac{{{m}_{\text{dd}}}.C%}{100}=\frac{14,7a}{100}=0,147a(g)$

${{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{m}_{{{H}_{2}}S{{O}_{4}}}}}{{{M}_{{{H}_{2}}S{{O}_{4}}}}}=\frac{0,147a}{98}$(mol)

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

 $\frac{a}{1000}$$\leftarrow $ $\frac{0,147a}{98}$ $\to $              $\frac{0,147a}{98}$  

Theo phương trình:

${{n}_{{{H}_{2}}}}={{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{0,147a}{98}$ ; ${{n}_{Al}}=\frac{2}{3}{{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{2}{3}.\frac{0,147a}{98}=\frac{a}{1000}$ (mol)

Khối lượng dung dịch sau phản ứng tăng 0,6g

$\Rightarrow {{m}_{\text{dd}\uparrow }}={{m}_{KL}}-{{m}_{{{H}_{2}}\uparrow }}=27.\frac{a}{1000}-2.\frac{0,147a}{98}=0,6g$

$\Rightarrow a=25$ 

                      Chọn đáp án C.