Đáp án: `V_{O_2}=2,24l`
Giải:
`n_{Fe}=\frac{16,8}{56}=0,3 \ (mol)`
`n_{SO_2}=\frac{5,6}{22,4}=0,25 \ (mol)`
$\mathop{Fe}\limits^0 → \mathop{Fe}\limits^{+3}+3e$
$\mathop{O}\limits^0+2e → \mathop{O}\limits^{-2}$
$\mathop{S}\limits^{+6}+2e → \mathop{S}\limits^{+4}$
Bảo toàn e:
`3n_{Fe}=2n_O+2n_{SO_2}`
→ `3.0,3=2n_O+2.0,25`
→ `n_O=0,2 \ (mol)`
→ `n_{O_2}=\frac{1}{2}n_O=\frac{1}{2}.0,2=0,1 \ (mol)`
→ `V_{O_2}=0,1.22,4=2,24 \ (l)`