$\text{ ta có : }$ $n_{CO}=0,2 mol$
$Fe_{x}$$O_{y}+yCO → xFe+$ $yCO_{2}$
$\text{ M = }$ $\dfrac{m_{hh}}{n_{hh}}=$ $\dfrac{44.ay+28.(0,2-ay)}{ay+0,2-ay}=40$
$\text{ → ay = 0,15 mol }$
$\text{ ta có : }$ $m_{Fe}$$_{xOy}= a ( 56x+16y)$
$\text{ = 56ax + 16ay = 8 gam }$
$\text{ → ax = 0,1 mol }$
$→\dfrac{x}{y}=$ $\dfrac{ax}{ay}=$ $\dfrac{0,1}{0,15}=$ $\dfrac{2}{3}$
$\text{ công thức o xit là }$ $Fe_{2}$$O_{3}$
$\text{ % }$$V_{CO}$$_{2}=$ $\dfrac{0,15}{0,2}.100=75 $$\text{ % }$