`n_{Fe}=m/M=7/{56}=0,125(mol)`
`Fe+2HCl->FeCl_2+H_2`
1 mol `Fe` 2 mol `HCl` 1 mol `FeCl_2` 1 mol `H_2`
`0,125` mol `Fe` y mol `HCl` z mol `FeCl_2` a mol `H_2`
Theo PTHH ta có:
`n_{FeCl_2}=n_{Fe}=z=0,125(mol)`
`⇒m_{FeCl_2}=0,125.127=15,875(g)`
`n_{H_2}=n_{Fe}=a=0,125(mol)`
`⇒m_{H_2}=0,125.2=0,25(g)`
`n_{HCl}=y=0,125.2=0,25(mol)`
`V=400ml=0,4l`
`⇒C_M=n/V=0,25:0,4=0,625(M)`
`⇒x=0,625(M)`