Đáp án đúng: B
450 ml
Ta có: $\displaystyle X+HCl\,\,\,\,\,\,\,\,\,\,\,\,\xrightarrow{BTE}{{n}_{Al}}=\frac{0,24.2}{3}=0,16\,mol$
$\displaystyle Y+NaOH\,\,\,\xrightarrow{BTE}n_{Al}^{du}=\frac{0,03.2}{3}=0,02\,\,mol$
$\displaystyle \to n_{Al}^{pu}=0,14\,mol\xrightarrow{BTNT.Al}{{n}_{A{{l}_{2}}{{O}_{3}}}}=0,07\,mol$
Vậy X có:
$\displaystyle \left\{ \begin{array}{l}{{n}_{Al}}=0,16\,mol\\{{n}_{F{{e}_{2}}{{O}_{3}}}}=0,07\,mol\end{array} \right.\,\,\,\,\xrightarrow{BTE+BTNT.H}\sum{{{n}_{{{H}^{+}}}}}=0,16.3+0,07.3.2=0,9\,mol$
$\displaystyle \to \sum{{{n}_{{{H}^{+}}}}}=1.V+0,5.2.V=0,9\,\to V=0,45\,lit$