Cho 1 lượng hỗn hợp bột Mg vào dung dịch X gồm FeC...

B

- Hướng dẫn giải

Phương pháp giải:

 $Mg + \left\{ \begin{gathered}

FeC{l_2} \hfill \\
CuC{l_2} \hfill \\
\end{gathered} \right. \to MgC{l_2}:0,15(mol)$

1: FeCl₂; CuCl₂ hết

$\begin{gathered}
Mg + C{u^{2 + }} \to M{g^{2 + }} + Cu \hfill \\
x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \hfill \\
Mg + F{e^{2 + }} \to M{g^{2 + }} + Fe \hfill \\
y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,y \hfill \\
x + y = 0,15 \hfill \\
\vartriangle m = 5,2(g) \hfill \\
\vartriangle {m_1} = 64x - 24x = 40{\text{x}} \hfill \\
\vartriangle {m_2} = 56y - 24y = 32y \hfill \\
\Rightarrow 40x + 32y = 5,2 \hfill \\
\Rightarrow x,y \hfill \\
\end{gathered} $

Giải chi tiết:

 $Mg + \left\{ \begin{gathered}

FeC{l_2} \hfill \\
CuC{l_2} \hfill \\
\end{gathered} \right. \to MgC{l_2}:0,15(mol)$

1: FeCl₂; CuCl₂ hết 

$\begin{gathered}
Mg + C{u^{2 + }} \to M{g^{2 + }} + Cu \hfill \\
x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x\,\,\,\,\,\,\,\,\,\,\,\,\,\,x \hfill \\
Mg + F{e^{2 + }} \to M{g^{2 + }} + Fe \hfill \\
y\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,\,\,\,y \hfill \\
x + y = 0,15 \hfill \\
\vartriangle m = 5,2(g) \hfill \\
\vartriangle {m_1} = 64x - 24x = 40{\text{x}} \hfill \\
\vartriangle {m_2} = 56y - 24y = 32y \hfill \\
\Rightarrow 40x + 32y = 5,2 \hfill \\
\Rightarrow \left\{ \begin{gathered}
x = 0,05 \hfill \\
y = 0,1 \hfill \\
\end{gathered} \right. \hfill \\
\end{gathered} $

=>  ${m_{CuC{l_2}}} = 0,05.135 = 6,75g$

Đáp án B