Question

When the supply of oxygen is limited, iron metal reacts with oxygen to produce a mixture of and . In a certain experiment, 20.00 g iron metal was reacted with 11.19 g oxygen gas. After the experiment, the iron was totally consumed, and 3.47 g oxygen gas remained. Calculate the amounts of and formed in this experiment?

Answer 1

Iron is reacting with O2 to form FeO and Fe2O3

Balanced reaction will be

3 Fe + 2 O2 ------> FeO + Fe2O3

Moles of Fe reacted = Mass / Atomic mass = 20 / 56 = 0.357

Moles of O2 reacted = mass / molar mass = 11.19-3.47 / 32 = 0.241 moles

The 0.357 moles of Fe will react with 0.24 moles of O2 to give

i) 0.357/ 3  moles of FeO = 0.119 moles of FeO

MAss of FeO = moles X molar mass = 0.119 X 71.84 = 8.55 grams

ii) 0.357/ 3  moles of Fe2O3 = 0.119 moles of Fe2O3

MAss of Fe2O3 = moles X molar mass = 0.119 X 160 = 19.04 grams