`a)` Khi cho hỗn hợp `X` gồm `Fe` và `Cu` vào dung dịch `HCl` thì `Cu` không tan

`⇒m_{Cu}=m_{cr}=6,4(g)`

`⇒%m_{Cu}=\frac{6,4.100%}{12}=53,33%`

`⇒%m_{Fe}=100%-53,33%=46,67%`

`b)m_{Fe}=12-6,4=5,6(g)`

`⇒n_{Fe}=\frac{5,6}{56}=0,1(mol)`

`Fe+2HCl→FeCl_{2}+H_{2}`

`0,1`     `0,2`                    `0,1`         / `mol`

`⇒V_{H_{2}}=0,1.22,4=2,24` ( lít )

`c)C_{%_{ddHCl}}=\frac{0,2.36,5}{200}.100%=3,65%`

`a)`

`Fe+2HCl->FeCl_2+H_2`

`->m_{Cu}=6,4(g)`

`->\%m_{Cu}={6,4}/{12}.100\%=53,33\%`

`->\%m_{Fe}=100-53,33=46,67\%`

`b)`

`n_{Fe}={12-6,4}/{56}=0,1(mol)`

`n_{H_2}=n_{Fe}=0,1(mol)`

`V_{H_2}=0,1.24=2,4(L)`

`c)`

`n_{HCl}=2n_{Fe}=0,2(mol)`

`C\%_{HCl}={0,2.36,5}/{200}.100\%=3,65\%`