Cho m gam Al tác dụng với $400 \mathrm{ml}$ dung dịch hỗn hợp...

Sau khi thêm $\mathrm{KOH}$ vào dung dịch $\mathrm{X}$ thu được 2 kết tủa nên $\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ dư. Kết tủa gồm $\mathrm{Al}(\mathrm{OH})_{3}$ và $\mathrm{Cu}(\mathrm{OH})_{2}$.
Hỗn hơp kim loại gồm: $\text{Ag,} \text{Cu}$
$n_{\mathrm{Cu}}=\dfrac{57,28-0,4 \cdot 108}{64}=0,22$
Vậy dung dịch X gồm: $\mathrm{X}\left\{\begin{array}{l}\mathrm{Al}^{3+} \\ \mathrm{Cu}^{2+} \\ \mathrm{NO}_{3}^{-}\end{array}\right.$
$57,28 \operatorname{gam} \mathrm{KL}\left\{\begin{array}{l}\mathrm{Ag}: 0,4 \\ \mathrm{Cu}: 0,22\end{array}\right.$
BT $\text{Cu}:{{n}_{\text{C}{{\text{u}}^{2+}}(\text{X})}}={{n}_{\text{Cu}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}\text{ (ban }\!\!\tilde{\mathrm{n}}\!\!\text{ a }\!\!\grave{\mathrm{a}}\!\!\text{ u)}}}-{{n}_{\text{Cu}}}=0,4x-0,22\to {{n}_{\text{Cu}{{(\text{OH})}_{2}}}}=0,4x-0,22$
BT $\text{e}:3{{n}_{\text{A}{{\text{l}}^{3+}}}}={{n}_{\text{Ag}}}+2{{n}_{\text{Cu}}}\to {{n}_{\text{A}{{\text{l}}^{3+}}}}=0,28$
$\mathrm{Cu}^{2+}+2 \mathrm{OH}^{-} \rightarrow \mathrm{Cu}(\mathrm{OH})_{2}$
$\mathrm{Al}^{3+}+3 \mathrm{OH}^{-} \rightarrow \mathrm{Al}(\mathrm{OH})_{3}$
$0,28$ $0,28 \quad 0,84$ $0,28$
$\mathrm{Al}(\mathrm{OH})_{3}+\mathrm{OH}^{-} \rightarrow \mathrm{AlO}_{2}^{-}+2 \mathrm{H}_{2} \mathrm{O}$
$y$ $y$
$\rightarrow\left\{\begin{array}{l}
98 .(0,4 x-0,22)+78 .(0,28-y)=27,37 \\
2 .(0,4 x-0,22)+0,84+y=1,225
\end{array} \rightarrow\left\{\begin{array}{l}
x=0,9 \\
y=0,105
\end{array}\right.\right.$