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

* Hướng dẫn giải

Đáp án B

 $\left\{\begin{array}{l}Al:x\\ Ca:y\\ C:\end{array}\right.\stackrel{H_{2}O}{\to }\left\{\begin{array}{l}\left\{\begin{array}{l}C{H}_4\\ C_2{H}_2\end{array}\right.\stackrel{O_2}{\to }\left\{\begin{array}{l}C{O}_2:0,25\\ H_{2}O:0,575\end{array}\right.\\ F:\left\{\begin{array}{l}Al{O}_2^{-}:x\\ O{H}^{-}\\ C{a}^{2+}:y\end{array}\right.\end{array}\right.$

BT  C:  $n_C=n_{C{O}_2}=0,25$

 $\begin{array}{l}Al\to A{l}^{3+}+3e{O}_2+4e\to 2O^{2-}\\ Ca\to C{a}^{2+}+2e\\ C\to \stackrel{+4}{C}+4e\end{array}$

BTNT O:  $n_{O_2}=\frac{2n_{C{O}_2}+H_{2}O}{2}=0,5375mol$

BT e:  $3x+2y+4.0,25=4.0,5375$

$$\begin{gathered} \begin{array}{l}27x+40y+12.0,25=17,75\\ \to \left\{\begin{array}{l}x=0,25\\ y=0,2\end{array}\right.\end{array} \\ F:\left\{\begin{array}{l}Al{O}_2^{-}:0,25\\ O{H}^{-}\\ C{a}^{2+}:0,2\end{array}\right.BTDT:n_{O{H}^{-}}=2.0,2-0,25=0,15 \\ \begin{array}{l}{H}^{+}+O{H}^{-}\to H_{2}O\\ 0,15\end{array} \\ \begin{array}{l}{H}^{+}+Al{O}_2^{-}+H_{2}O\to Al{\left(OH\right)}_3\left(1\right)\\ 0,250,25\end{array} \\ Al{\left(OH\right)}_3+3H^{+}\to A{l}^{3+}+3H_{2}O\left(2\right) \\ \begin{array}{l}{n}_{H^{+}\left(2\right)}=0,5-0,15-0,25=0,1\\ n_{Al{\left(OH\right)}_3}=n_{Al{\left(OH\right)}_3\left(1\right)}-n_{Al{\left(OH\right)}_3\left(2\right)}=\frac{13}{60}mol\\ m_{Al{\left(OH\right)}_3}=16,9gam\end{array} \end{gathered}$$