a.FexOy+yCO->xFe+yCO2
Gọi x là nCO pư.
Ta có:Mhh khí=20*2=40
nCO=4,48/22,4=0,2(mol)
40=\(\dfrac{28\cdot\left(0,2-x\right)+44x}{0,2-x+x}\)=>x=0,15
=>nCO pư=0,15(mol)=>nFexOy=0,15/y(mol)
=>MFexOy=\(\dfrac{8}{\dfrac{0,15}{y}}\)
Mặt khác ta có:56x+16y=\(\dfrac{8}{\dfrac{0,15}{y}}\)
=>\(\dfrac{x}{y}=\dfrac{2}{3}\) Vậy CT: Fe2O3
b.%VCO2=%nCO2=\(\dfrac{0,15\cdot100}{0,2}=75\%\)