Đáp án C
$\left\{ \begin{array}{l} Na:0,1{\rm{ mol}}\\ {\rm{Al}}:3{\rm{ gam}} \end{array} \right.$$\xrightarrow{{{H}_{2}}O}$$\left\{ \begin{array}{l} N{a^ + }:0,1{\rm{ mol}}\left( {{n_{N{a^ + }}} = {n_{Na}}} \right)\\ AlO_2^ - :0,1{\rm{ mol}}\left( {BT{\rm{D}}T:{n_{AlO_2^ - }} = {n_{N{a^ + }}}} \right)\\ Al{\rm{ du}} \end{array} \right.$
${{n}_{Al\text{ phan ung}}}={{n}_{AlO_{2}^{-}}}=0,1$mol
BTNT: ${{m}_{Al\text{ du}}}={{m}_{Al\text{ ban dau}}}-{{m}_{Al\text{ phan ung}}}=3-0,1.27=0,3$(g)
