Hòa tan hoàn toàn hỗn hợp Al và Al2O3 trong 200ml dung dịch HCl...

$\left\{ \begin{aligned}
& Al \\
& A{{l}_{2}}{{O}_{3}} \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& Al \\
& O \\
\end{aligned} \right.\xrightarrow{+HCl} $dung dịch X $ \left\{ \begin{aligned}
& AlC{{l}_{3}} :x \\
& HCl \text{d} \text{:y} \\
& {{\text{H}}_{2}}O \\
\end{aligned} \right.$
Yêu cầu bài toán: $a=\dfrac{3x+y}{0,2}$
Khi cho NaOH vào X ta có các công thức cần nhớ
$\left\{ \begin{aligned}
& \text{*}\text{B }\!\!{\scriptscriptstyle 3\!/\!{ }_4}\!\!\text{ t u xu }\!\!\hat{\mathrm{E}}\!\!\text{ t hi }\!\!\ddot{\mathrm{O}}\!\!\text{ n k }\!\!\tilde{\mathrm{O}}\!\!\text{ t t }\!\!\tilde{\mathrm{n}}\!\!\text{ a: HCl d v }\!\!\tilde{\mathrm{o}}\!\!\text{ a h }\!\!\tilde{\mathrm{O}}\!\!\text{ t} \\
& \text{*}\text{Khi Al(OH}{{\text{)}}_{3}}\text{cha tan l }\!\!{}^\text{1}\!\!\text{ i:} Al{{(OH)}_{3}}=\dfrac{1}{3}(O{{H}^{-}}-{{H}^{+}}) \\
& *\text{Khi Al(OH}{{\text{)}}_{3}}\text{ }\!\!\cdot\!\!\text{ tan l }\!\!{}^\text{1}\!\!\text{ i:} \text{Al(OH}{{\text{)}}_{3}}=4A{{l}^{3+}}-(O{{H}^{-}}-{{H}^{+}}) \\
\end{aligned} \right.$
$\begin{aligned}
& \Leftrightarrow \left\{ \begin{aligned}
& y=0,1 \\
& \dfrac{m}{78}=\dfrac{1}{3}(0,25-0,1) \\
& \dfrac{m}{78}=4x-(0,45-0,1) \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=0,1 \\
& m=3,9 gam \\
& x=0,1 \\
\end{aligned} \right. \\
& \Rightarrow a=\dfrac{3.0,1+0,1}{0,2}=2M \\
\end{aligned}$