a) 39,32% + 25,54% + 28,07% = 92,92%
=> Sai đề
b) CTHH: \(Fe_xS_yO_z\)
Ta có: \(\%Fe=\frac{56x}{152}.100\%=36,84\%\) => x = 1 (mol)
\(\%S=\frac{32y}{152}.100\%=21,05\%\) => y = 1(mol)
\(\%O=\frac{16z}{152}.100\%=42,11\%\) => z = 4 (mol)
=> CTHH: \(FeSO_4\)