Đáp án đúng: C
Cho V (lít) dung dịch AlCl3 1,0M vào V (lít) dung dịch NaAlO2 1M.
$\begin{array}{l}{HCl}\,{+}\,{NaAl}{{{O}}_{{2}}}\,{+}\,{{{H}}_{{2}}}{O}\xrightarrow{{}}\,{Al(OH}{{{)}}_{{3}}}\,{+}\,{NaCl}\\\,\,{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{{}}\,\,\,\,\,\,\,\,\,\,\,{1}\\{3HCl}\,{+}\,{Al(OH}{{{)}}_{{3}}}\,\xrightarrow{{}}\,{AlC}{{{l}}_{{3}}}\,{+}\,{3}{{{H}}_{{2}}}{O}\\\,\,\,\,{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{1}}{{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{1}}{{3}}\\{NaOH}\,{+}\,{AlC}{{{l}}_{{3}}}\,\xrightarrow{{}}\,{Al(OH}{{{)}}_{{3}}}\,{+}\,{NaCl}\\\,\,\,\,\,{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{1}}{{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{1}}{{3}}\\{A}{{{l}}^{{3+}}}\,{+}\,\,{3AlO}_{{2}}^{{-}}\,{+}\,{6}{{{H}}_{{2}}}{O}\,\xrightarrow{{}}\,\,{4Al(OH}{{{)}}_{{3}}}\\\frac{{1}}{{3}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{4}}{{3}}\end{array}$