Đáp án B
$m\left\{ \begin{array}{l} Ba\\ Na\\ A{l_2}{O_3} \end{array} \right.$$\xrightarrow{{{H}_{2}}O}$$\left\{ \begin{array}{l} {H_2}:0,095{\rm{ }}mol\\ A{l_2}{O_3}:0,318m \end{array} \right.$
$m\left\{ \begin{array}{l} Ba\\ Na\\ A{l_2}{O_3} \end{array} \right.$$\xrightarrow{{{H}_{2}}S{{O}_{4}}}11,65\text{ }gam\text{ }BaS{{O}_{4}}:0,05$
${{n}_{Ba}}={{n}_{BaS{{O}_{4}}}}=0,05$
BT e: $2{{n}_{{{H}_{2}}}}=2{{n}_{Ba}}+{{n}_{Na}}\to {{n}_{Na}}=0,09$
Vì sau phản ứng Al$_{2}$O$_{3}$ dư nên dung dịch sau phản ứng chứa:
$\left\{ \begin{array}{l} Ba{\left( {Al{O_2}} \right)_2}:0,05\left( {{n_{Ba}}} \right)\\ NaAl{O_2}:0,09\left( {{n_{Na}}} \right) \end{array} \right.$
BT Al: $2{{n}_{A{{l}_{2}}{{O}_{3}}\left( \text{pu} \right)}}=2{{n}_{Ba{{\left( Al{{O}_{2}} \right)}_{2}}+{{n}_{NaAl{{O}_{2}}}}}}=0,19$
$\to {{n}_{A{{l}_{2}}{{O}_{3}}\left( \text{pu} \right)}}=0,095$
$\left\{ \begin{array}{l} {M_{X\left( {phan\,ung} \right)}} = m - 0,318m = 0,682m\\ {m_{X\left( {phan\,ung} \right)}} = {m_{Ba}} + {m_{Na}} + {m_{A{l_2}{O_3}\left( {phan\,ung} \right)}} = 18,61 \end{array} \right.$
$\to m=27,29\left( g \right)$.
