Đáp án đúng: D
7 : 4.
Ta có : $\displaystyle \left\{ \begin{array}{l}{{{n}}_{{O}{{{H}}^{{-}}}}}{=0,612(mol)}\\{{{n}}_{{Al(OH}{{{)}}_{{3}}}}}{=0,108(mol)}\end{array} \right.$
=> Kết tủa đã bị tan 1 phần.
Trong 0,4 lít E sẽ có:$\displaystyle \left\{ \begin{array}{l}{{{n}}_{{A}{{{l}}^{{3+}}}}}{=0,4(x+2y)}\\{{{n}}_{{SO}_{{4}}^{{2-}}}}{=0,4}{.3}{.y}\xrightarrow{{BaC}{{{l}}_{{2}}}}{1,2}{.y=}{{{n}}_{\downarrow }}{=0,144}\Rightarrow {y=0,12}\end{array} \right.$
Ta lại có:$\displaystyle {{{n}}_{{A}{{{l}}^{{3+}}}}}{=0,4(x+2y)}\xrightarrow{{BTNT}{.Al}}\left\{ \begin{array}{l}{Al(OH}{{{)}}_{{3}}}{:0,108}\\{AlO}_{{2}}^{{-}}{:0,4(x+2y)-0,108}\end{array} \right.$
Vậy trong F có:$\displaystyle \left\{ \begin{array}{l}{AlO}_{{2}}^{{-}}{:0,4(x+2}{.0,12)-0,108}\,\,\,\,\,\,\,\,\,{(mol)}\\{C}{{{l}}^{{-}}}{:0,4}{.3x(mol)}\\{SO}_{{4}}^{{2-}}{:0,4}{.3}{.0,12(mol)}\\{N}{{{a}}^{{+}}}{:0,612(mol)}\end{array} \right.$
$\displaystyle \xrightarrow{{BTDT}}{0,4(x+2}{.0,12)-0,108+0,4}{.3x+2}{.0,4}{.3}{.0,12=0,612}$
⇒ x = 0,21 mol
