Hỗn hợp E có khối lượng 17,75 gam gồm Al, Ca, Al4C3 và CaC2. Hòa...

$\left\{ \begin{aligned}
& Al:x \\
& Ca:y \\
& C: \\
\end{aligned} \right.\xrightarrow{{{H}_{2}}O}\left\{ \begin{aligned}
& \left\{ \begin{aligned}
& C{{H}_{4}} \\
& {{C}_{2}}{{H}_{2}} \\
\end{aligned} \right.\xrightarrow{{{O}_{2}}}\left\{ \begin{aligned}
& C{{O}_{2}}:0,25 \\
& {{H}_{2}}O:0,575 \\
\end{aligned} \right. \\
& F:\left\{ \begin{aligned}
& AlO_{2}^{-}:x \\
& O{{H}^{-}} \\
& C{{a}^{2+}}:y \\
\end{aligned} \right. \\
\end{aligned} \right.$
BT C: ${{n}_{C}}={{n}_{C{{O}_{2}}}}=0,25$
$\begin{aligned}
& Al\to A{{l}^{3+}}+3e {{O}_{2}}+4e\to 2{{O}^{2-}} \\
& Ca\to C{{a}^{2+}}+2e \\
& C\to \overset{+4}{\mathop{C}} +4e \\
\end{aligned}$
BTNT O: ${{n}_{{{O}_{2}}}}=\dfrac{2{{n}_{C{{O}_{2}}}}+{{H}_{2}}O}{2}=0,5375 mol$
BT e: $3x+2y+4.0,25=4.0,5375$
$\begin{aligned}
& 27x+40y+12.0,25=17,75 \\
& \to \left\{ \begin{aligned}
& x=0,25 \\
& y=0,2 \\
\end{aligned} \right. \\
\end{aligned}$
$F:\left\{ \begin{aligned}
& AlO_{2}^{-}:0,25 \\
& O{{H}^{-}} \\
& C{{a}^{2+}}:0,2 \\
\end{aligned} \right.BTDT: {{n}_{O{{H}^{-}}}}=2.0,2-0,25=0,15$
$\begin{aligned}
& {{H}^{+}}+O{{H}^{-}}\to {{H}_{2}}O \\
& 0,15 \\
\end{aligned}$
$\begin{aligned}
& {{H}^{+}}+AlO_{2}^{-}+{{H}_{2}}O\to Al{{(OH)}_{3}} (1) \\
& 0,25 0,25 \\
\end{aligned}$
$Al{{\left( OH \right)}_{3}}+3{{H}^{+}}\to A{{l}^{3+}}+3{{H}_{2}}O (2)$
$\begin{aligned}
& {{n}_{{{H}^{+}}(2)}}=0,5-0,15-0,25=0,1 \\
& {{n}_{Al{{(OH)}_{3}}}}={{n}_{Al{{(OH)}_{3}}(1)}}-{{n}_{Al{{(OH)}_{3}}\left( 2 \right)}}=\dfrac{13}{60} mol \\
& {{m}_{Al{{(OH)}_{3}}}}=16,9 gam \\
\end{aligned}$
Tính nhanh: ${{n}_{Al{{(OH)}_{3}}}}=\dfrac{4{{n}_{AlO_{2}^{-}}}-{{n}_{{{H}^{+}}}}+{{n}_{O{{H}^{-}}}}}{3}$