`n_{CuO}=x(mol);n_{Fe_2O_3}=y(mol)`

`m_B=80x+160y=19,6(1)`

`CuO+H_2` $\xrightarrow{t^o}$ `Cu+H_2O`

`Fe_2O_3+3H_2` $\xrightarrow{t^o}$ `2Fe+3H_2O`

`n_{H_2\ l t}=x+3y={6,72}/{22,4}=0,3(2)`

`(1)(2)->x=0,135;y=0,055`

`n_{CuO\  t t}=0,135.60\%=0,081(mol)`

`n_{Fe_2O_3\   t t}=0,055.60\%=0,033(mol)`

`->n_{Cu}=0,081(mol);n_{Fe}=0,033.2=0,066(mol)`

Vậy trong X

`m_{Cu}=0,081.64=5,184(g)`

`m_{Fe}=0,066.56=3,696(g)`

`m_{CuO}=(0,135-0,081).80=4,32(g)`

`m_{Fe_2O_3}=(0,055-0,033).160=3,52(g)`