$Fe_xO_y+ 2yHCl\to xFeCl_{\frac{2y}{x}}+ yH_2O$
$n_{Fe_xO_y}=\dfrac{14,4}{56x+16y}$
$\Rightarrow n_{\text{muối}}=\dfrac{14,4x}{56x+16y}$
$\Rightarrow M_{\text{muối}}=\dfrac{25,4(56x+16y)}{14,4x}=\dfrac{1422,4x+406,4y}{14,4x}=56+35,5\dfrac{2y}{x}$
$\Leftrightarrow 1422,4x+406,4y=14,4(56x+35,5.2y)$
$\Leftrightarrow \dfrac{x}{y}=1$
$\Rightarrow x=y=1$
Vậy oxit là $FeO$