Cho m gam hỗn hợp X gồm Al, Al2O3, Al(OH)3 (trong đó oxi chiếm 47,625% khối

* Hướng dẫn giải

Đáp án D.

$$\begin{gathered} m\text{\hspace{0.17em}X}\left\{\begin{array}{l}\text{Al}\\ \begin{array}{l}{\text{Al}}_2{\text{O}}_3:a\\ \text{Al}{\left(OH\right)}_3:b\end{array}\end{array}\stackrel{HCl}{\to }\right.\left\{\begin{array}{l}{\text{H}}_2:0,18\\ {\text{AlCl}}_3\stackrel{\text{NaOH}:1,6848}{\to }\left\{\text{Al}{\left(\text{OH}\right)}_3:0,3952\right.\end{array}\right. \\ {n}_{\text{Al}}=\frac{2}{3}{n}_{{\text{H}}_2}=0,12 \end{gathered}$$

$n_{\text{NaOH}}>3n_{\text{Al}{\left(\text{OH}\right)}_3}\to$Kết tủa tạo thành đã tan 1 phần:

 $$\begin{gathered} {\text{Al}}^{3+}+3{\text{OH}}^{-}\to \text{Al}{\left(\text{OH}\right)}_3 \\ \text{x 3x x} \\ \text{Al}{\left(\text{OH}\right)}_3+{\text{OH}}^{-}\to {\text{AlO}}_2^{-}+2{\text{H}}_2\text{O} \\ y y \\ \left\{\begin{array}{l}x-y=0,3952\\ 3x+y=1,6848\end{array}\right.\to \left\{\begin{array}{l}x=0,52\\ y=0,1248\end{array}\right. \\ \to n_{{\text{Al}}^{3+}}=0,52 \end{gathered}$$

BT $\text{Al}:0,12+2a+b=0,52$

Oxi chiếm 47, 265% khối lượng nên:

$$\begin{gathered} \frac{(3a+3b)\cdot 16}{27.0,12+102a+78b}\cdot 100=47,265 \\ \to \left\{\begin{array}{l}a=0,13\\ b=0,14\end{array}\right. \\ \to \%m_{{\text{Al}}_2{\text{O}}_3}=48,36\% \end{gathered}$$