`n_{Fe}=\frac{16,8}{56}=0,3(mol)`
`n_{SO_2}=\frac{4,48}{22,4}=0,2(mol)`
Coi hỗn hợp `X` gồm `0,3(mol)` `Fe` và `x(mol)` `O`
Sơ đồ cho-nhận e
\(\rm \mathop{Fe}\limits^{+0}\to \mathop{Fe}\limits^{+3}+3e \\\mathop{O}\limits^{+0}+2e\to \mathop{O}\limits^{-2} \\\mathop{S}\limits^{+6}+2e\to \mathop{S}\limits^{+4}\)
BTe
`->3n_{Fe}=2n_{O}+2n_{SO_2}`
`->n_{O}=\frac{0,9-0,4}{2}=0,25(mol)`
`->m_{O}=0,25.16=4(g)`
`->m=m_{Fe}+m_{O}=16,8+4=20,8(g)`