Cho m gam Al tác dụng với 400 ml dung dịch hỗn hợp \(\mathrm{AgNO}_{3} 1 \mathrm{M}\) và \(\mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) xM thu được dung dịch X và 57,28 gam hỗn hợ...

* Hướng dẫn giải

Đáp án B.

Sau khi thêm \(\mathrm{KOH}\) vào dung dịch 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: \(\text{Ag,}\,\text{Cu}\)

\(n_{\mathrm{Cu}}=\frac{57,28-0,4 \cdot 108}{64}=0,22\)

Vậy dung dịch X gồm: \(\mathrm{X}\left\{\begin{array}{l}\mathrm{Al}^{3+} \\ \mathrm{Cu}^{2+} \\ \mathrm{NO}_{3}^{-}\end{array}\right.\)

\(m\,\text{Al}\frac{\text{AgN}{{\text{O}}_{3}}:0,4}{\text{Cu}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}:0,4x}\left\{ \begin{align} & \text{X}\left\{ \begin{array}{*{35}{l}} \text{A}{{\text{l}}^{3+}} \\ \text{C}{{\text{u}}^{2+}} \\ \text{NO}_{3}^{-} \\ \end{array} \right.\to \left\{ \downarrow 27,37\left\{ \begin{array}{*{35}{l}} \text{Al}{{(\text{OH})}_{3}} \\ \text{Cu}{{(\text{OH})}_{2}} \\ \end{array} \right. \right. \\ & 57,28\operatorname{gam}\text{KL}\left\{ \begin{array}{*{35}{l}} \text{Ag}:0,4 \\ \text{Cu}:0,22 \\ \end{array} \right. \\ \end{align} \right.\)

BT \(\text{Cu}:{{n}_{\text{C}{{\text{u}}^{2+}}(\text{X})}}={{n}_{\text{Cu}{{\left( \text{N}{{\text{O}}_{3}} \right)}_{2}}\text{ (ban dau)}}}-{{n}_{\text{Cu}}}=0,4x-0,22\to {{n}_{\text{Cu}{{(\text{OH})}_{2}}}}=0,4x-0,22\)

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

\(\rightarrow\left\{\begin{array}{l} 98 .(0,4 x-0,22)+78 .(0,28-y)=27,37 \\ 2 .(0,4 x-0,22)+0,84+y=1,225 \end{array} \rightarrow\left\{\begin{array}{l} x=0,9 \\ y=0,105 \end{array}\right.\right.\)