\(n_{Fe3O4}=\frac{41,76}{232}=0,18\left(mol\right)\)
\(Fe_3O_4+8HCl\rightarrow2FeCl_3+FeCl_2+4H_2O\)
\(\Rightarrow n_{FeCl3}=0,36\left(mol\right)\)
\(Cu+2FeCl_3\rightarrow CuCl_2+2FeCl_2\)
\(\Rightarrow n_{Cu}=0,18\left(mol\right)\)
\(\Rightarrow m=0,18.64=11,52\left(g\right)\)