Cho dung dịch X gồm Al2(SO4)3 và 0,12 mol H2SO4 vào dung dịch chứa

* Hướng dẫn giải

Đáp án B.

$$\begin{gathered} X\left\{\begin{array}{l}{\text{Al}}_2{\left({\text{SO}}_4\right)}_3:0,08\\ {\text{H}}_2{\text{SO}}_4:0,12\end{array}\stackrel{\text{Ba}{\left(\text{OH}\right)}_2:0,4}{\to }\right.\left\{\begin{array}{l}\begin{array}{l}{\text{BaSO}}_4\\ \text{Al}{\left(OH\right)}_3\end{array}\end{array}\stackrel{{\text{t}}^0}{\to }\left\{\begin{array}{l}{\text{Al}}_2{\text{O}}_3\\ {\text{BaSO}}_4\end{array}\right.\right. \\ \begin{array}{l}{\text{H}}^{+}+{\text{OH}}^{-}\to {\text{H}}_2\text{O}\\ 0,240,8\end{array} \\ {\text{Al}}^{3+}+3{\text{OH}}^{-}\to \text{Al}{\left(\text{OH}\right)}_3 \\ \text{0,16 0,48 0,16} \\ \text{Al}{\left(\text{OH}\right)}_3+{\text{OH}}^{-}\to {\text{AlO}}_2^{-}+2{\text{H}}_2\text{O} \\ 0,08\leftarrow (0,8-0,24-0,48) \\ {n}_{\text{Al}{\left(\text{OH}\right)}_3}=0,16-0,08=0,08 \\ \to n_{{\text{Al}}_2{\text{O}}_3}=0,04 \\ {\text{Ba}}^{2+}+{\text{SO}}_4^{2-}\to {\text{BaSO}}_4 \\ 0,4(0,24+0,12)\to 0,36 \\ m=m_{{\text{Al}}_2{\text{O}}_3}+m_{{\text{BaSO}}_4}=0,04.102+0,36.233=87,96( \text{g}) \end{gathered}$$