\(\begin{array}{l}
Fe+2HCl\to FeCl_2+H_2\\
Theo\,PT:\,n_{Fe}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)\\
2Fe+3Cl_2\xrightarrow{t^o}2FeCl_3\\
Cu+Cl_2\xrightarrow{t^o}CuCl_2\\
Theo\,PT:\,n_{FeCl_3}=n_{Fe}=0,2(mol)\\
\to n_{Cu}=n_{CuCl_2}=\frac{48,7-0,2.162,5}{135}=0,12(mol)\\
\to m=0,12.64+0,2.56=18,88(g)\\
Theo\,PT:\,n_{HCl}=2n_{H_2}=0,4(mol)\\
\to V_{dd\,HCl(dung)}=\dfrac{\frac{0,4}{2}}{100\%-10\%}\approx 0,22(l)
\end{array}\)