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

* Hướng dẫn giải

$+\left\{\begin{array}{l}Al\\ Ba\\ O\end{array}\right\}\stackrel{H_{2}O}{\to }Y\left\{\begin{array}{l}Ba{\left(OH\right)}_2:xmol\\ Ba{\left(Al{O}_2\right)}_2:ymol\end{array}\right\};\left\{\begin{array}{l}TN1:Y\stackrel{\mathrm{0,27}molC{O}_2}{\to }\left\{\begin{array}{l}Al{\left(OH\right)}_3\downarrow \\ BaC{O}_3\downarrow \end{array}\right\}+Ba{\left(HC{O}_3\right)}_2\\ TN1:Y\stackrel{C{O}_{2}dư}{\to } Al{\left(OH\right)}_3\downarrow +Ba{\left(HC{O}_3\right)}_2\end{array}\right.$

$\begin{array}{l}Các phản ứng tạo kết tủa:\left\{\begin{array}{l}Ba{\left(OH\right)}_2+C{O}_2\stackrel{}{\to }BaC{O}_3+H_{2}O\\ Ba{\left(Al{O}_2\right)}_2+2C{O}_2+4H_{2}O\stackrel{}{\to }Ba{\left(HC{O}_3\right)}_2+2Al\downarrow \end{array}\right.\\ \Rightarrow \left\{\begin{array}{l}TN1:197a+78.2y=\mathrm{21,51}\\ TN2:78.2y=\mathrm{15,6}\end{array}\right.\Rightarrow \left\{\begin{array}{l}a=\mathrm{0,03}\\ y=\mathrm{0,1}\end{array}\right.\Rightarrow n_{Ba}=\mathrm{0,03}+\frac{\mathrm{0,27}-\mathrm{0,03}}{2}=\mathrm{0,15}\\ \Rightarrow BTE:n_O=\frac{3n_{Al}+2n_{Ba}-2n_{H_2}}{2}=\frac{\mathrm{0,2.3}+\mathrm{0,15.2}-\mathrm{0,2.2}}{2}=\mathrm{0,25}mol\Rightarrow \boxed{m_X=\mathrm{29,95}gam}\end{array}$