Hướng dẫn giải:

$\begin{align}  & {{n}_{KOH}}=0,14mol,{{n}_{AlC{{l}_{3}}}}=0,04mol \\  & A{{l}^{3+}}+3O{{H}^{-}}\to Al{{(OH)}_{3}} \\  & 0,04\text{      0,12         0,04    mol} \\ & Al{{(OH)}_{3}}+O{{H}^{-}}\to Al{{O}_{2}}^{-}+{{H}_{2}}O \\  & 0,02\text{            0,02          0,02mol} \\  & \text{m}\downarrow \text{=0,02}\text{.78=1,56g}\text{.} \\\end{align}$

Chọn A.