`Cu+HCl->X`
`Fe+2HCl->FeCl_2+H_2`
`FeCl_2+2NaOH->Fe(OH)_2+2NaCl`
`a)`$Ket\ tua:Fe(OH)_2$
`n_{Fe(OH)_2}=(18)/(56+16.2+2)=0,2(mol)`
`n_{FeCl_2}=n_{Fe(OH)_2}=0,2(mol)`
`n_{Fe}=n_{FeCl_2}=0,2(mol)`
`%m_{Fe}=(0,2.56)/(17,6%)=63,64%`
`%m_{Cu}=100%-63,64%=36,36%`
`b)``n_{HCl}=2.0,2=0,4(mol)`
`mdd_{HCl}=(0,4.36,5)/(6,63%)=220,2111614(g)`
`Vdd=(220,2111614)/(1,1)=200,1919649(ml)`