Đáp án đúng: B
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