Đặt \(\left\{{}\begin{matrix}x=n_{Fe\left(OH\right)_3}\left(mol\right)\\y=n_{Cu\left(OH\right)_2}\left(mol\right)\end{matrix}\right.\)
\(PTHH:2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\\ Cu\left(OH\right)_2+H_2SO_4\rightarrow CuSO_4+2H_2O\\ m_{CT_{H_2SO_4}}=\dfrac{400\cdot9,8\%}{100\%}=39,2\left(g\right)\\ \Rightarrow n_{H_2SO_4}=\dfrac{39,2}{98}=0,4\left(mol\right)\)
Theo đề ta có HPT: \(\left\{{}\begin{matrix}107x+98y=31,2\\1,5x+y=0,4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,2\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(\Rightarrow m_{Fe\left(OH\right)_3}=0,2\cdot107=21,4\left(g\right)\\ \Rightarrow\left\{{}\begin{matrix}\%_{Fe\left(OH\right)_3}=\dfrac{21,4}{31,2}\cdot100\%\approx69\%\\\%_{Cu\left(OH\right)_2}=100\%-69\%=31\%\end{matrix}\right.\)
