a) \(Fe+2HCL\rightarrow FeCl_2+H_2\uparrow\)
Cu không tác dụng được với HCL
a, PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)
b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
Theo PT: \(n_{Fe}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Fe}=0,2.56=11,2\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{Fe}=\dfrac{11,2}{20}.100\%=56\%\\\%m_{Cu}=100-56=44\%\end{matrix}\right.\)
c, Theo PT: \(n_{HCl}=2n_{H_2}=0,4\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
\(\Rightarrow C\%_{ddHCl}=\dfrac{14,6}{300}.100\%\approx4,87\%\)
Bạn tham khảo nhé!