A 7,65 gam
B 6,75 gam
C 5,76 gam
D 5,67gam
B
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} $
$Mg + \left\{ \begin{gathered}
FeC{l_2} \hfill \\
CuC{l_2} \hfill \\
\end{gathered} \right. \to MgC{l_2}:0,15(mol)$
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} $
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