Lời giải:\(m\left\{ \begin{array}{l}
Ba\\
Na\\
A{l_2}{O_3}
\end{array} \right. \to \left\{ \begin{array}{l}
{H_2}:0,095{\rm{ }}mol\\
A{l_2}{O_3}:0,318m
\end{array} \right.\)
\(m\left\{ \begin{array}{l}
Ba\\
Na\\
A{l_2}{O_3}
\end{array} \right. \to 11,65{\rm{ }}gam{\rm{ }}BaS{O_4}:0,05\)
\({{n}_{Ba}}={{n}_{BaS{{O}_{4}}}}=0,05\)
BT e: \(2{{n}_{{{H}_{2}}}}=2{{n}_{Ba}}+{{n}_{Na}}\to {{n}_{Na}}=0,09\)
Vì sau phản ứng Al2O3 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: \(2{n_{A{l_2}{O_3}\left( {{\rm{phan ung}}} \right)}} = 2{n_{Ba{{\left( {Al{O_2}} \right)}_2} + {n_{NaAl{O_2}}}}} = 0,19\)
\( \to {n_{A{l_2}{O_3}\left( {{\rm{phan ung}}} \right)}} = 0,095\)
\(\left\{ \begin{array}{l}
{M_{X\left( {{\rm{phan ung}}} \right)}} = m - 0,318m = 0,682m\\
{m_{X\left( {{\rm{phan ung}}} \right)}} = {m_{Ba}} + {m_{Na}} + {m_{A{l_2}{O_3}\left( {{\rm{phan ung}}} \right)}} = 18,61
\end{array} \right.\)
→ m = 27,29 gam.
