Quy đổi hỗn hợp $X$: $Fe$ ($x$ mol), $O$ ($y$ mol)
$\to 56x+16y=10,44$ $(1)$
$n_{NO_2}=\dfrac{8,97}{46}=0,195(mol)$
$\mathop{Fe}\limits^{0}\to \mathop{Fe}\limits^{+3}+3e$
$\mathop{O}\limits^0+2e\to \mathop{O}\limits^{-2}$
$\mathop{N}\limits^{+5}+1e\to \mathop{N}\limits^{+4}$
Bảo toàn e: $3n_{Fe}=2n_O+n_{NO_2}$
$\to 3x-2y=0,195$ $(2)$
$(1)(2)\to x=0,15; y=0,1275$
Bảo toàn $Fe$: $n_{Fe_2O_3\rm bđ}=\dfrac{n_{Fe}}{2}=0,075(mol)$
$\to m=0,075.160=12g$