a) CTHH: SxOy
%S= \(\frac{32x}{80}.100\%=40\%\) => x = 1(mol)
%O = \(\frac{16y}{80}.100\%=100\%-40\%\) => y = 3(mol)
=> CTHH: SO3
b) CTHH: FexOy
%Fe = \(\frac{56x}{160}.100\%=70\%\) => x = 2 (mol)
%O = \(\frac{16y}{160}.100\%=100\%-70\%\) => y = 3 (mol)
=> CTHH: Fe2O3