1.
$n_{AgCl}=\dfrac{22,6}{143,5}=0,157 mol$
$FeCl_x+xAgNO_3\to Fe(NO_3)_x+xAgCl$
$\Rightarrow n_{FeCl_x}=\dfrac{0,157}{x}$
$M_{FeCl_x}=\dfrac{10x}{0,157}=63,69x=56+35,5x$
$\Leftrightarrow x=1,98\approx 2$
Vậy muối là $FeCl_2$
2.
$n_{H_2SO_4}=\dfrac{7,84}{98}=0,08 mol$
$2R_xO_y+2yH_2SO_4\to xR_2(SO_4)_{\frac{2y}{x}}+ 2yH_2O$
$\Rightarrow n_{R_xO_y}=\dfrac{0,08}{y}$
$M_{R_xO_y}=\dfrac{4,48y}{0,08}=\dfrac{56y}{x}$
$\Rightarrow x=y=1; R=56(Fe)$
Vậy oxit là $FeO$