3g chất rắn không tan là Cu
=> \(m_{Zn}+m_{Fe}=18,6\left(g\right)\)
Theo đề bài ta có PTHH:
\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\) (I)
\(Fe+H_2SO_4\rightarrow FeSO_4+H_2\) (II)
Gọi \(n_{Zn}=x\left(mol\right);n_{Fe}=y\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}65x+56y=18,6\\x+y=0,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
\(\Rightarrow m_{Fe}=56\cdot0,1=5,6\left(g\right)\) ; \(m_{Zn}=65\cdot0,2=13\left(g\right)\)
\(\%m_{Fe}=\dfrac{5,6}{21,6}\cdot100\%=25,93\left(\%\right)\)
\(\%m_{Zn}=\dfrac{13}{21,6}\cdot100\%=60,19\left(\%\right)\)
\(\%m_{Cu}=\left(100-25,93-60,19\right)\%=13,88\%\)
