Chất rắn không tan là Fe
\(2Fe+3Cl_2\rightarrow2FeCl_3\)
0,1 0,1
=>mFe=5,6(g)
=>\(m_{Cu}+m_{Mg}=14.4-5.6=9.8\left(g\right)\)
Đặt \(n_{Cu}=x;n_{Mg}=y\)
Ta có hệ:
\(\left\{{}\begin{matrix}2x+2y=2\cdot0.2=0.4\\64x+24y=9.8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0.125\\y=0.075\end{matrix}\right.\)
\(\%m_{Fe}=\dfrac{5.6}{14.4}=38.89\%\)
\(\%m_{Cu}=\dfrac{0.125\cdot64}{14.4}=55.56\%\)
\(\%m_{Mg}=1-38.89\%-55.56\%=5.55\%\)