- Chọn D.

- Ta có: ${{n}_{Al}}=\frac{4,05}{27}=0,15\left( mol \right)$.

- PTHH: $4Al+3{{O}_{2}}\to 2A{{l}_{2}}{{O}_{3}}$

      0,15(mol) $\to 0,1125\left( mol \right)$

$2KMn{{O}_{4}}\xrightarrow{{{t}^{o}}}{{K}_{2}}Mn{{O}_{4}}+Mn{{O}_{2}}+{{O}_{2}}$

 $0,25\left( mol \right)\leftarrow $                               0,1125(mol)

- khối lượng $KMn{{O}_{4}}$cần dùng là: $m=0,25\times 158=39,5\left( gam \right)$.