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Câu hỏi: Cho dung dịch $\mathrm{X}$ gồm $0,08~\text{mol} \text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}$ và $0,12~ \text{mol} {{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ vào dung dịch chứa $0,4~\text{mol} \text{Ba}{{(\text{OH})}_{2}}$ thu được kết tủa $\mathrm{Y}$. Đem nung nóng kết tủa $\mathrm{Y}$ ngoài không khí đến khối lượng không đổi thu được m gam chất rắn khan. Giá trị của m là
A. $90,12.$
B. $87,96.$
C. $91,86$.
D. $92,45.$
A. $90,12.$
B. $87,96.$
C. $91,86$.
D. $92,45.$
$X\left\{ \begin{array}{*{35}{l}}
\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}:0,08 \\
{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}:0,12 \\
\end{array}\xrightarrow{\text{Ba}{{(\text{OH})}_{2}}:0,4} \right.\left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& \text{BaS}{{\text{O}}_{4}} \\
& \text{Al}{{(\text{OH})}_{3}} \\
\end{aligned} \\
\end{array}\xrightarrow{{{\text{t}}^{0}}}\left\{ \begin{array}{*{35}{l}}
\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}} \\
\text{BaS}{{\text{O}}_{4}} \\
\end{array} \right. \right.$
$\begin{aligned}
& {{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}\to {{\text{H}}_{2}}\text{O} \\
& 0,24 0,8 \\
\end{aligned}$
$\mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}$
$0,16\quad 0,48\quad 0,16$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \rightarrow \mathrm{AlO}_{2}^{-}+2 \mathrm{H}_{2} \mathrm{O}$
$0,08 \leftarrow(0,8-0,24-0,48)$
${{n}_{\text{Al}{{(\text{OH})}_{3}}}}=0,16-0,08=0,08$
$\rightarrow n_{\mathrm{Al}_{2} \mathrm{O}_{3}}=0,04$
$\mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-} \rightarrow \mathrm{BaSO}_{4}$
$0,4(0,24+0,12) \rightarrow 0,36$
$m={{m}_{\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}}}+{{m}_{\text{BaS}{{\text{O}}_{4}}}}=0,04.102+0,36.233=87,96 (~\text{g})$
\text{A}{{\text{l}}_{2}}{{\left( \text{S}{{\text{O}}_{4}} \right)}_{3}}:0,08 \\
{{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}:0,12 \\
\end{array}\xrightarrow{\text{Ba}{{(\text{OH})}_{2}}:0,4} \right.\left\{ \begin{array}{*{35}{l}}
\begin{aligned}
& \text{BaS}{{\text{O}}_{4}} \\
& \text{Al}{{(\text{OH})}_{3}} \\
\end{aligned} \\
\end{array}\xrightarrow{{{\text{t}}^{0}}}\left\{ \begin{array}{*{35}{l}}
\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}} \\
\text{BaS}{{\text{O}}_{4}} \\
\end{array} \right. \right.$
$\begin{aligned}
& {{\text{H}}^{+}}+\text{O}{{\text{H}}^{-}}\to {{\text{H}}_{2}}\text{O} \\
& 0,24 0,8 \\
\end{aligned}$
$\mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}$
$0,16\quad 0,48\quad 0,16$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \rightarrow \mathrm{AlO}_{2}^{-}+2 \mathrm{H}_{2} \mathrm{O}$
$0,08 \leftarrow(0,8-0,24-0,48)$
${{n}_{\text{Al}{{(\text{OH})}_{3}}}}=0,16-0,08=0,08$
$\rightarrow n_{\mathrm{Al}_{2} \mathrm{O}_{3}}=0,04$
$\mathrm{Ba}^{2+}+\mathrm{SO}_{4}^{2-} \rightarrow \mathrm{BaSO}_{4}$
$0,4(0,24+0,12) \rightarrow 0,36$
$m={{m}_{\text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}}}+{{m}_{\text{BaS}{{\text{O}}_{4}}}}=0,04.102+0,36.233=87,96 (~\text{g})$
Đáp án B.