Cu không tác dụng với dd HCl.
PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
a, Ta có: \(n_{H_2}=\frac{0,448}{22,4}=0,02\left(mol\right)\)
Theo PT: \(n_{HCl}=2n_{H_2}=0,04\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,04.36,5=1,46\left(g\right)\)
\(\Rightarrow m_{ddHCl}=\frac{1,46.100\%}{3,65\%}=40\left(g\right)\)
b, Theo PT: \(n_{Fe}=n_{H_2}=0,02\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,02.56=1,12\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\frac{1,12}{7}.100\%=16\%\\\%m_{Cu}=100-16=84\%\end{matrix}\right.\)
Bạn tham khảo nhé!