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Câu hỏi: Cho m gam hỗn hợp $\mathrm{X}$ gồm $\mathrm{Al}, \mathrm{Al}_{2} \mathrm{O}_{3}, \mathrm{Al}(\mathrm{OH})_{3}$ (trong đó oxi chiếm $47,265 \%$ khối lượng) tác dụng với dung dịch HCl vừa đủ thu được dung dịch $X$ và 4,032 lít $H_{2}$ (đktc). Cho $842,4 \mathrm{ml}$ dung dịch $\text{NaOH} 2\text{M}$ vào dung dịch $\mathrm{X}$ thu được 30,8256 gam kết tủa. Phần trăm khối lượng $\mathrm{Al}_{2} \mathrm{O}_{3}$ trong hỗn hợp $\mathrm{X}$ là
A. $45,84 \%$.
B. $46,72 \%$.
C. $47,18 \%$.
D. $48,36 \%$.
A. $45,84 \%$.
B. $46,72 \%$.
C. $47,18 \%$.
D. $48,36 \%$.
$m \text{X}\left\{ \begin{array}{*{35}{l}}
\text{Al} \\
\begin{aligned}
& \text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}:a \\
& \text{Al}{{(\text{OH})}_{3}}:b \\
\end{aligned} \\
\end{array}\xrightarrow{HCl} \right.\left\{ \begin{array}{*{35}{l}}
{{\text{H}}_{2}}:0,18 \\
\text{AlC}{{\text{l}}_{3}}\xrightarrow{\text{NaOH}:1,6848}\left\{ \text{Al}{{(\text{OH})}_{3}}:0,3952 \right. \\
\end{array} \right.$
$n_{\mathrm{Al}}=\dfrac{2}{3} n_{\mathrm{H}_{2}}=0,12$
$n_{\mathrm{NaOH}}>3 n_{\mathrm{Al}(\mathrm{OH})_{3}} \rightarrow$ Kết tủa tạo thành đã tan 1 phần:
$\mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}$
$x$ $x \quad 3 x$ $x$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \rightarrow \mathrm{AlO}_{2}^{-}+2 \mathrm{H}_{2} \mathrm{O}$
$y$ $y$
$\left\{ \begin{array}{*{35}{l}}
x-y=0,3952 \\
3x+y=1,6848 \\
\end{array} \right.$$\to \left\{ \begin{array}{*{35}{l}}
x=0,52 \\
y=0,1248 \\
\end{array} \right.$
$\rightarrow n_{\mathrm{Al}^{3+}}=0,52$
BT $\text{Al}: 0,12+2a+b=0,52$
Oxi chiếm $47,265 \%$ khối lượng nên:
$\dfrac{(3 a+3 b) \cdot 16}{27.0,12+102 a+78 b} \cdot 100=47,265$
$\rightarrow\left\{\begin{array}{l}a=0,13 \\ b=0,14\end{array}\right.$
$\rightarrow \% m_{\mathrm{Al}_{2} \mathrm{O}_{3}}=48,36 \%$
\text{Al} \\
\begin{aligned}
& \text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}:a \\
& \text{Al}{{(\text{OH})}_{3}}:b \\
\end{aligned} \\
\end{array}\xrightarrow{HCl} \right.\left\{ \begin{array}{*{35}{l}}
{{\text{H}}_{2}}:0,18 \\
\text{AlC}{{\text{l}}_{3}}\xrightarrow{\text{NaOH}:1,6848}\left\{ \text{Al}{{(\text{OH})}_{3}}:0,3952 \right. \\
\end{array} \right.$
$n_{\mathrm{Al}}=\dfrac{2}{3} n_{\mathrm{H}_{2}}=0,12$
$n_{\mathrm{NaOH}}>3 n_{\mathrm{Al}(\mathrm{OH})_{3}} \rightarrow$ Kết tủa tạo thành đã tan 1 phần:
$\mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}$
$x$ $x \quad 3 x$ $x$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \rightarrow \mathrm{AlO}_{2}^{-}+2 \mathrm{H}_{2} \mathrm{O}$
$y$ $y$
$\left\{ \begin{array}{*{35}{l}}
x-y=0,3952 \\
3x+y=1,6848 \\
\end{array} \right.$$\to \left\{ \begin{array}{*{35}{l}}
x=0,52 \\
y=0,1248 \\
\end{array} \right.$
$\rightarrow n_{\mathrm{Al}^{3+}}=0,52$
BT $\text{Al}: 0,12+2a+b=0,52$
Oxi chiếm $47,265 \%$ khối lượng nên:
$\dfrac{(3 a+3 b) \cdot 16}{27.0,12+102 a+78 b} \cdot 100=47,265$
$\rightarrow\left\{\begin{array}{l}a=0,13 \\ b=0,14\end{array}\right.$
$\rightarrow \% m_{\mathrm{Al}_{2} \mathrm{O}_{3}}=48,36 \%$
Đáp án D.