PTHH: \(8HCl+Fe_3O_4\rightarrow FeCl_2+2FeCl_3+4H_2O\)
a) Ta có: \(\left\{{}\begin{matrix}n_{Fe_3O_4}=\frac{23,2}{232}=0,1\left(mol\right)\\n_{HCl}=\frac{200\cdot3,65\%}{36,5}=0,2\left(mol\right)\end{matrix}\right.\)
Xét tỉ lệ: \(\frac{0,2}{8}< \frac{0,1}{1}\) \(\Rightarrow\) HCl phản ứng hết, Fe3O4 còn dư
\(\Rightarrow n_{Fe_3O_4\left(pư\right)}=0,025\left(mol\right)\) \(\Rightarrow n_{Fe_3O_4\left(dư\right)}=0,07mol\) \(\Rightarrow m_{Fe_3O_4}=0,075\cdot232=17,4\left(g\right)\)
b) Theo PTHH: \(\left\{{}\begin{matrix}n_{FeCl_2}=\frac{1}{8}n_{HCl}=0,025mol\\n_{FeCl_3}=\frac{1}{4}n_{HCl}=0,05mol\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}m_{FeCl_2}=0,025\cdot127=3,175\left(g\right)\\m_{FeCl_3}=0,05\cdot162,5=8,125\left(g\right)\end{matrix}\right.\)
c) Ta có: \(m_{dd}=m_{Fe_3O_4}+m_{ddHCl}-m_{Fe_3O_4\left(dư\right)}=23,2+200-17,4=205,8\left(g\right)\)
\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\frac{3,175}{205,8}\cdot100\approx1,54\%\\C\%_{FeCl_3}=\frac{8,125}{205,8}\cdot100\approx3,95\%\end{matrix}\right.\)
