nHCl = 0.3 x 0.5 = 0.15 mol
nO(oxit) = 1/2nHCl = 0.075 mol
-> mO(oxit) = 0.075 x 16 = 1.2g
-> mFe (oxit/0 = 4 - 1.2 = 2.8g
-> nFe (oxit) = 0.05 mol
FexOy thì x : y = 0.05 : 0.075 = 2 : 3
=> Fe2O3.
d A/H2 = 17 => M A = 34
CO2: 44 6
34
CO : 28 10
=> nCO2/nCO = 6/10 = 3/5
=>% V CO2 = 3/(3+5) = 37.5%
=> % V CO = 100% - 37.5% = 62.5%