A. 45,84 %
B. 46,72 %
C. 47,18 %
D. 48,36 %
D
Đáp án D.
\(m\,\text{X}\left\{ \begin{array}{*{35}{l}} \text{Al} \\ \begin{align} & \text{A}{{\text{l}}_{2}}{{\text{O}}_{3}}:a \\ & \text{Al}{{(\text{OH})}_{3}}:b \\ \end{align} \\ \end{array}\to \right.\left\{ \begin{array}{*{35}{l}} {{\text{H}}_{2}}:0,18 \\ \text{AlC}{{\text{l}}_{3}}\to \left\{ \text{Al}{{(\text{OH})}_{3}}:0,3952 \right. \\ \end{array} \right.\)
\(n_{\mathrm{Al}}=\frac{2}{3} n_{\mathrm{H}_{2}}=0,12\)
\(n_{\mathrm{NaOH}}>3 n_{\mathrm{Al}(\mathrm{OH})_{3}} \rightarrow\) 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:
\(\frac{(3 a+3 b) \cdot 16}{27.0,12+102 a+78 b} \cdot 100=47,265\)
\(\rightarrow\left\{\begin{array}{l}a=0,13 \\ b=0,14\end{array}\right.\)
\(\rightarrow \% m_{\mathrm{Al}_{2} \mathrm{O}_{3}}=48,36 \%\)
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