Đáp án:

`O x it: Fe_2O_3`

Giải thích các bước giải:

`Pt :CuO+H_2` $\xrightarrow{t^o}$ `Cu+H_2O\ (1)`

       `Fe_xO_y+yH_2` $\xrightarrow{t^o}$ `xFe+yH_2O\ (2)`

       `Fe+2HCl->FeCl_2+H_2\ (3)`

`n_{H_2}={0,448}/{22,4}=0,02(mol)`

`n_{Fe(3)}=n_{H_2}=0,02(mol)`

`n_{Cu}={1,76-(0,02.56)}/{64}=0,01(mol)`

`n_{CuO(1)}=n_{Cu}=0,01(mol)`

`m_{Fe_xO_y}=2,4-(0,01.80)=1,6(g)`

`n_{Fe_2O_y(2)}=1/{x}.n_{Fe}={0,02}/{x}(mol)`

`M_{Fe_xO_y}=1,6:{0,02}/{x}=80x(g//mol)`

`<=>56x+16y=80x`

`<=>16y=24x`

`=>x/y={16}/{24}=2/3`

`Chon\ x=2\ ; y=3`

`O x it: Fe_2O_3`

\(\begin{array}{l}
Dat\,oxit\,la\,Fe_xO_y\\
Fe+2HCl\to FeCl_2+H_2\\
Theo\,PT:\,n_{Fe}=n_{H_2}=\frac{0,448}{22,4}=0,02(mol)\\
\to n_{Cu}=\frac{1,76-0,02.56}{64}=0,01(mol)\\
CuO+H_2\xrightarrow{t^o}Cu+H_2O\\
Fe_xO_y+yH_2\xrightarrow{t^o}xFe+yH_2O\\
Theo\,PT:\,n_{CuO}=n_{Cu}=0,01(mol)\\
\to m_{Fe_xO_y}=2,4-0,01.80=1,6(g)\\
Theo\,PT:\,n_{Fe_xO_y}.x=n_{Fe}\\
\to \frac{1,6x}{56x+16y}=0,02\\
\to \frac{x}{y}=\frac{2}{3}\\
\to Oxit:\,Fe_2O_3
\end{array}\)