Week 5 · Chapter 1 (E): pure iron and its +2 and +3 states, the reactions of iron (air, steam, chlorine, sulphur, acids, passivity) and the three oxides of iron with their preparation and properties (paper equations and calculations only) · 3 lessons
2
Iron(II) oxide and iron(III) oxide
Chemistry Sec 3 Term 1, Chapter 1: pp. 20–24
GoalCompare iron(II) oxide and iron(III) oxide: colour, preparation, reactions with acids and with oxygen, and calculate masses.
New words
iron(II) oxide · FeO, a black powder that does not dissolve in wateriron(III) oxide · Fe₂O₃, a reddish-brown solid, found as the ore hematite and used as a red pigmentiron(III) hydroxide · Fe(OH)₃, a reddish-brown solid (a precipitate) formed when ammonia solution meets an iron(III) salt solution
Explanation
Iron(II) oxide FeO is a black powder that is insoluble in water. On paper it is prepared by reducing a higher oxide, for example Fe₂O₃ + CO → 2FeO + CO₂, or by heating iron(II) oxalate without air: FeC₂O₄ → FeO + CO + CO₂. In hot air it oxidises to the red oxide, 4FeO + O₂ → 2Fe₂O₃. With acids it gives an iron(II) salt and water: FeO + H₂SO₄ → FeSO₄ + H₂O. Iron(III) oxide Fe₂O₃ is reddish-brown. On paper, ammonia solution added to an iron(III) salt gives a reddish-brown precipitate of Fe(OH)₃, and heating that gives the oxide: FeCl₃ + 3NH₄OH → Fe(OH)₃ + 3NH₄Cl, then 2Fe(OH)₃ → Fe₂O₃ + 3H₂O. A second route is heating iron(II) sulphate, 2FeSO₄ → Fe₂O₃ + SO₂ + SO₃ (the gases are poisonous, so this stays on paper). With hot concentrated acids Fe₂O₃ gives iron(III) salts: Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O. The rule is: oxide + acid → salt + water, and the iron keeps its state, so FeO gives iron(II) salts and Fe₂O₃ gives iron(III) salts.
FeO
Fe₂O₃
colour
black
reddish-brown
iron state
+2
+3
found as
—
hematite
with acid
iron(II) salt + water
iron(III) salt + water
in hot air
turns into Fe₂O₃
no change
Worked examples
ƒEquation
Fe₂O₃(s) + 3H₂SO₄(aq) → Fe₂(SO₄)₃(aq) + 3H₂O(l)
ratio 1 : 3 : 1 : 3
Fe₂O₃ with sulphuric acid (Fe = 56, S = 32, O = 16, H = 1): Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O. For 16 g of Fe₂O₃, n = 16 ÷ 160 = 0.1 mol. Acid: 3 × 0.1 = 0.3 mol × 98 = 29.4 g. Salt: 0.1 mol × 400 g/mol = 40.0 g. Water: 0.3 mol × 18 = 5.4 g. Check: 16 + 29.4 = 45.4 g and 40 + 5.4 = 45.4 g.
Heating the hydroxide on paper: 2Fe(OH)₃ → Fe₂O₃ + 3H₂O. M(Fe(OH)₃) = 56 + 3 × 17 = 107 g/mol, so 21.4 g is 0.2 mol. The ratio 2 : 1 : 3 gives 0.1 mol of Fe₂O₃ = 16 g and 0.3 mol of water = 5.4 g, and 16 + 5.4 = 21.4 g.
Oxide
Equation with H₂SO₄
Moles of acid per mole of oxide
FeO
FeO + H₂SO₄ → FeSO₄ + H₂O
1
Fe₂O₃
Fe₂O₃ + 3H₂SO₄ → Fe₂(SO₄)₃ + 3H₂O
3
The same acid, different amounts. One FeO needs one H₂SO₄, but one Fe₂O₃ needs three, because it holds three oxygen atoms that must leave as water.
Class activity
A colour chart on paper. With crayons, make a chart of the colours in the lesson: FeO black, Fe₂O₃ reddish-brown, Fe(OH)₃ reddish-brown precipitate. If you did the rusting test, add the colour of the rust on your nail and say which oxide it looks like. Then write on a card, for FeO and Fe₂O₃, the one sentence “oxide + acid gives …” with the right state of iron.
Practice
1
FeO + H₂SO₄ → FeSO₄ + H₂O. For 14.4 g of FeO (Fe = 56, S = 32, O = 16, H = 1) find the mass of acid needed and of FeSO₄ formed.
