Lời giải:

 $\begin{align} & {{n}_{AlC{{l}_{3}}}}=0,1\text{ }mol,\text{ }{{\text{n}}_{NaOH}}=0,45mol. \\  & A{{l}^{3+}}+3O{{H}^{-}}\to Al{{(OH)}_{3}} \\ & 0,1\text{        0,3            0,1  mol} \\  & Al{{(OH)}_{3}}+O{{H}^{-}}\to Al{{O}_{2}}^{-}+{{H}_{2}}O \\ & 0,1\text{                0,1       0,1    mol} \\  & \text{NaOH  + C}{{\text{O}}_{2}}\text{  }\to \text{  NaHC}{{\text{O}}_{3}} \\  & 0,05\text{           0,05  mol} \\  & Al{{O}_{2}}^{-}+\text{C}{{\text{O}}_{2}}\text{ }+2{{H}_{2}}O\to Al{{(OH)}_{3}}+\text{HC}{{\text{O}}_{3}}^{-} \\  & 0,1\text{         0,1} \\  & {{\text{n}}_{C{{O}_{2}}}}=0,15mol \\ \end{align}$

=>Chọn đáp án A