Fe3O4 + 8HCl → 2FeCl3 + FeCl2 + 4H2O (1)
x<----------8x<-----------2x<--------x
2FeCl3 + Cu → 2FeCl2 + CuCl2 (2)
2x---------->x------->2x
\(Fe^{2+}+2OH^-\rightarrow Fe\left(OH\right)_2\)
3x-------->6x------------->3x
\(Cu^{2+}+2OH^-\rightarrow Cu\left(OH\right)_2\)
x---------->2x-------------->x
\(H^++OH^-\rightarrow H_2O\)
\(n_{OH^-}=0,5.1+0,5.1=1\left(mol\right)\)
\(m_{KLkhongtan}=m_{Cu}\)
Dung dịch Y : FeCl2, FeCl3, HCl dư
Theo PT (1) \(n_{FeCl_2}=x\left(mol\right);n_{FeCl_3}=2x\left(mol\right)\)
Theo PT (2) \(n_{CuCl_2}=x\left(mol\right);n_{FeCl_2}=2x\left(mol\right)\)
Theo PT (2) : \(\Sigma n_{FeCl_2}=3x\left(mol\right)\)
Sau phản ứng => 36,8 gam kết tủa
\(m_{kt}=m_{Cu\left(OH\right)_2}+m_{Fe\left(OH\right)_2}=x.98+3x.90=36,8\)
=> x=0,1(mol)
\(m=m_{Cu\left(pứ\right)}+m_{Cu\left(dư\right)}+m_{Fe_3O_4}=0,1.64+1,6+0,1.232=31,2\left(g\right)\)
\(n_{HCl\left(dư\right)}=1-\left(2x+6x\right)=0,2\left(mol\right)\)
\(n_{HCl\left(pứ\right)}=8x=0,8\left(mol\right)\)
=> \(n_{HCl\left(bđ\right)}=0,8+0,2=1\left(mol\right)\)
