PT: \(Fe\left(OH\right)_2+2HCl\rightarrow FeCl_2+2H_2O\)
\(Cu\left(OH\right)_2+2HCl\rightarrow CuCl_2+2H_2O\)
Ta có: 90nFe(OH)2 + 98nCu(OH)2 = 59,5 (1)
Theo PT: \(n_{H_2O}=2n_{Fe\left(OH\right)_2}+2n_{Cu\left(OH\right)_2}=\dfrac{22,68}{18}=1,26\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Fe\left(OH\right)_2}=0,28\left(mol\right)\\n_{Cu\left(OH\right)_2}=0,35\left(mol\right)\end{matrix}\right.\)
Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe\left(OH\right)_2}=0,28\left(mol\right)\\n_{CuCl_2}=n_{Cu\left(OH\right)_2}=0,35\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=n_{H_2O}=1,26\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{1,26.36,5}{15\%}=306,6\left(g\right)\)
Ta có: m dd sau pư = 59,5 + 306,6 = 366,1 (g)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,28.127}{366,1}.100\%\approx9,71\%\\C\%_{CuCl_2}=\dfrac{0,35.135}{366,1}.100\%\approx12,91\%\end{matrix}\right.\)
