$1/$
$a/$
$\%K=\frac{39.100\%}{122,5}=31,83\%$
$\%Cl=\frac{35,5.100\%}{122,5}=25,98\%$
$\%O=100\%-31,82\%-25,98\%=42,2\%$
$b/$
$\%Fe=\frac{56.2.100\%}{400}=20\%$
$\%S=\frac{32.3.100\%}{400}=24\%$
$\%O=100\%-20\%-24\%=56\%$
$2/$
$\%O=100\%-28,57\%-14,29\%=57,14\%$
$m_{O}=\frac{84.57,14}{100}=48g$
$→n_{O}=\frac{48}{16}=3mol$
$m_{Mg}=\frac{28,57.84}{100}=24g$
$→n_{Mg}=\frac{24}{24}=1mol$
$m_{C}=\frac{14,29.84}{100}=12g$
$→n_{C}=\frac{12}{12}=1mol$
Vậy $CTHH$ của hợp chất là $MgCO_{3}$