\(\left\{{}\begin{matrix}64.n_{Cu}+232.n_{Fe_3O_4}=36\\n_{Cu}:n_{Fe_3O_4}=2:1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_{Cu}=0,2\left(mol\right)\\n_{Fe_3O_4}=0,1\left(mol\right)\end{matrix}\right.\)
\(n_{HCl}=\dfrac{800.7,3\%}{36,5}=1,6\left(mol\right)\)
PTHH: \(Fe_3O_4+8HCl\rightarrow FeCl_2+2FeCl_3+4H_2O\)
0,1---->0,8------>0,1------->0,2
\(Cu+2FeCl_3\rightarrow CuCl_2+2FeCl_2\)
0,1<----0,2-------->0,1------>0,2
=> dd sau pư chứa \(\left\{{}\begin{matrix}FeCl_2:0,3\left(mol\right)\\CuCl_2:0,1\left(mol\right)\\HCl:0,8\left(mol\right)\end{matrix}\right.\)
mdd sau pư = 0,1.232 + 0,1.64 + 800 = 829,6 (g)
\(\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,3.127}{829,6}.100\%=4,59\%\\C\%_{CuCl_2}=\dfrac{0,1.135}{829,6}.100\%=1,63\%\\C\%_{HCl}=\dfrac{0,8.36,5}{829,6}.100\%=3,52\%\end{matrix}\right.\)
