Hòa tan hoàn toàn m gam hỗn hợp X gồm Ba; BaO, Al và AL2O3 vào nước dư,

* Hướng dẫn giải

Đáp án D.

Dựa vào đồ thị thấy một đoạn không có kết tủa chứng minh trong dung dịch Y còn ${\text{OH}}^{\text{-}}{\text{, AlO}}_2^{-},{\text{Ba}}^{2+}$

$$\begin{gathered} \begin{array}{l}{\text{H}}^{+}+{\text{OH}}^{-}\to {\text{H}}_2\text{O}\\ \text{0,2\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}0,2}\end{array} \\ \begin{array}{l}{\text{H}}^{+}+{\text{AlO}}_2^{-}+{\text{H}}_2\text{O}\to \text{Al}{\left(OH\right)}_3\\ xxx\end{array} \\ 3{\text{H}}^{+}+\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Al}{\left(\text{OH}\right)}_3\to {\text{Al}}^{3+}+3{\text{H}}_2\text{O} \\ (3x-0,6)\leftarrow (x-0,2) \\ \to 0,06.26+0,04.2=m_{{\text{C}}_2{\text{H}}_4\left(\text{Y}\right)}+m_{{\text{C}}_2{\text{H}}_2\left(\text{Y}\right)} \\ \to m_{{\text{C}}_2{\text{H}}_4\left(\text{Y}\right)}+m_{{\text{C}}_2{\text{H}}_2\left(\text{Y}\right)}=1,32\text{\hspace{0.17em}\hspace{0.17em}gam} \\ \to \text{X}\left\{\begin{array}{l}\text{Ba}:0,25\\ \text{Al}:0,3\to \text{BT\hspace{0.17em}\hspace{0.17em}e}:n_{\text{O}}=\frac{2n_{\text{Ba}}+3n_{\text{Al}}-2n_{{\text{H}}_2}}{2}=0,45\\ \text{O}\end{array}\right. \\ \to m=49,55 \end{gathered}$$