Coi $Fe_3O_4=FeO.Fe_2O_3$
X gồm $Fe$ (x mol) $FeO$ (y mol) $Fe_2O_3$ (z mol).
$\Rightarrow x=0,04$ (1)
Bảo toàn Fe:
$n_{FeCl_2}=n_{Fe}+n_{FeO}==x+y$
$n_{FeCl_3}=2n_{Fe_2O_3}=2z$
$\Rightarrow 127x+127y+162,5.2z=45,76$ (2)
$\Delta m=1,68g=m_{O_2}$
$\Rightarrow n_{O_2}=\dfrac{1,68}{32}=0,0525 mol$
Bảo toàn e: $3x+y=0,0525.4=0,21$ (3)
(1)(2)(3)$\Rightarrow x=0,04; y=z=0,09$
$\Rightarrow m=m_{Fe}+m_{FeO}+m_{Fe_2O_3}=56.0,04+72.0,09+160.0,09=23,12g$
$\%m_{Fe}=\dfrac{0,04.56.100}{23,12}=9,69\%$
$\%m_{FeO}=\dfrac{0,09.72.100}{23,12}=28,03\%$
$\%m_{Fe_2O_3}=62,28\%$