Hòa tan hoàn toàn m gam hỗn hợp X gồm Ba, BaO, Al và Al2O3 vào...

Từ đồ thị: $\to {{n}_{O{{H}^{-}}/Y}}={{0,2}^{mol}}$
$X\left\{ \begin{aligned}
& Ba:a\left( mol \right) \\
& Al:b\left( mol \right) \\
& O:c\left( mol \right) \\
\end{aligned} \right.\xrightarrow{{{H}_{2}}O}\left\langle \begin{aligned}
& Y\left\{ \begin{aligned}
& B{{a}^{2+}}:a\left( mol \right) \\
& AlO_{2}^{-}:b\left( mol \right) \\
& O{{H}^{-}}:0,2\left( mol \right) \\
\end{aligned} \right. \\
& {{H}_{2}}:0,25\left( mol \right) \\
\end{aligned} \right.$
$\left\{ \begin{aligned}
& BTe:2{{n}_{Ba}}+3{{n}_{Al}}=2{{n}_{O}}+2{{n}_{{{H}_{2}}}} \\
& BTDT:2{{n}_{B{{a}^{2+}}}}={{n}_{AlO_{2}^{-}}}+{{n}_{O{{H}^{-}}}} \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& 2a+3b=2c+0,5 \left( 1 \right) \\
& 2a=b+0,2 \left( 2 \right) \\
\end{aligned} \right.$
$Y\xrightarrow{HCl:1,4\left( mol \right)}\left\{ \begin{aligned}
& B{{a}^{2+}}:a\left( mol \right) \\
& A{{l}^{3+}}:b\left( mol \right)+{{H}_{2}}O \\
& C{{l}^{-}}:1,4\left( mol \right) \\
\end{aligned} \right.$
$BTDT:2a+3b=1,4$ (3)
$\xrightarrow{\left( 1 \right)\left( 2 \right)\left( 3 \right)}\left\{ \begin{aligned}
& a=0,25 \\
& b=0,3 \\
& c=0,45 \\
\end{aligned} \right.\to {{m}_{X}}={{m}_{Ba}}+{{m}_{Al}}+{{m}_{O}}=49,55\left( g \right)$