- Giả sử oxit là $Fe_3O_4$ (tạo 2 muối)
$n_{Fe_3O_4}=\dfrac{16}{232}=\dfrac{2}{29}$
$Fe_3O_4+8HCl\to 2FeCl_3+FeCl_2+4H_2O$
$\Rightarrow n_{FeCl_2}=\dfrac{2}{29}; n_{FeCl_3}=\dfrac{4}{29}$
$m_{\text{muối}}=\dfrac{2}{29}.127+\dfrac{4}{29}.162,5=31,17g$ (loại)
- Vậy oxit sắt có dạng $Fe_2O_x$
$Fe_2O_x+2xHCl\to 2FeCl_x+xH_2O$
$n_{Fe_2O_x}=\dfrac{16}{112+16x}$
$\Rightarrow n_{FeCl_x}=\dfrac{32}{112+16x}=\dfrac{32,5}{56+35,5x}$
$\Rightarrow x=3(Fe_2O_3)$
$n_{HCl}=6n_{Fe_2O_3}=0,6 mol$
$\Rightarrow C_{M_{HCl}}=\dfrac{0,6}{0,3}=2M$