- Phần 1: Gọi \(\left\{{}\begin{matrix}n_{CuCl_2}=x\left(mol\right)\\n_{MgCl_2}=y\left(mol\right)\end{matrix}\right.\)
Ta có: \(n_{AgCl}=\dfrac{14,35}{143,5}=0,1\left(mol\right)\)
PTHH:
\(CuCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Cu\left(NO_3\right)_2\)
x-------------------------->2x
\(\)\(MgCl_2+2AgNO_3\rightarrow2AgCl\downarrow+Mg\left(NO_3\right)_2\)
y-------------------------->2y
`=> 2x + 2y = 0,1 (1)`
- Phần 2:
\(CuCl_2+2NaOH\rightarrow Cu\left(OH\right)_2\downarrow+2NaCl\)
x-------------------------->x
\(MgCl_2+2NaOH\rightarrow Mg\left(OH\right)_2\downarrow+2NaCl\)
y-------------------------->y
\(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)
x--------------->x
\(Mg\left(OH\right)_2\xrightarrow[]{t^o}MgO+H_2O\)
y-------------->y
`=> 80x + 40y = 3,2 (2)`
Từ `(1), (2) => x = 0,03; y = 0,02`
`=>` \(\left\{{}\begin{matrix}\%m_{CuCl_2}=\dfrac{0,03.95}{0,03.95+0,02.135}=51,35\%\\\%m_{MgCl_2}=100\%-51,35\%=48,65\%\end{matrix}\right.\)
