\( Đặt:a=n_{CuO}\left(mol\right);b=n_{Fe_xO_y}\left(mol\right)\left(a,b>0\right)\\ CuO+H_2\rightarrow\left(t^o\right)Cu+H_2O\\ Fe_xO_y+yH_2\rightarrow xFe+yH_2O\\ Fe+2HCl\rightarrow FeCl_2+H_2\\ n_{Fe}=n_{H_2}=\dfrac{0,448}{22,4}=0,02\left(mol\right)\\ \Rightarrow m_{Cu}=1,76-0,02.56=0,64\left(g\right)\\ \Rightarrow n_{CuO}=n_{Cu}=\dfrac{0,64}{64}=0,01\left(mol\right)\\ \Rightarrow m_{CuO}=0,01.80=0,8\left(g\right)\\ \Rightarrow n_{Fe_xO_y}=2,4-0,8=1,6\left(g\right)\\ n_{O\left(trong.oxit.sắt\right)}=\dfrac{1,6-0,02.56}{16}=0,03\\ \Rightarrow x:y=0,02:0,03=2:3\\ \Rightarrow x=2;y=3\\ \Rightarrow CTHH:Fe_2O_3\)
