- TH1: oxit là $Fe_3O_4$
$n_{Fe_3O_4}=\dfrac{0,8}{232}=0,0034 mol$
$Fe_3O_4+8HCl\to FeCl_2+2FeCl_3+4H_2O$
$\Rightarrow n_{FeCl_2}=0,0034; n_{FeCl_3}=0,0068 mol$
$\Rightarrow m_{\text{muối}}=0,0034.127+0,0068.162,5$ (loại)
- TH2: oxit có dạng $R_2O_x$
$R_2O_x+2xHCl\to 2RCl_x+xH_2O$
$n_{R_2O_x}=\dfrac{0,8}{2R+16x}$
$\Rightarrow n_{RCl_x}=\dfrac{1,6}{2R+16x}$
$\Rightarrow \dfrac{1,6}{2R+16x}=\dfrac{1,9}{R+35,5x}$
$\Rightarrow 1,9(2R+16x)=1,6(R+35,5x)$
$\Rightarrow R=12x$
$x=2\Rightarrow R=24(Mg)$
Vậy oxit là $MgO$ (magie oxit)
