9/Fe + H2SO4 => FeSO4 + H2
....0.01.................................0.01
=> mFe =0.56 g
FeO + H2 => Fe + H2O
a.............................a
Fe2O3 + 3H2 =>2 Fe + 3H20
b.....................................3b
=> a + 3b = 0.05
72a + 160b = 4 - 0.56.2 =2.88
=> a=0.011; b=0.0128 => kq
10/Viết pthh
đặt nFe(FeO) = a; nFe(Fe2O3) = b
=> a + b = 0.013
a.160.1/2 + b.232.1/3 = 1.016
=> a,b => %
mO2 = m oxit - m Fe = 0.288 => nO2 =0.009 mol => V O2 =....