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Iron ores, extraction, steel and alloys

الأسبوع 4 · الباب الأول (٤): الحديد في الطبيعة وخاماته وتجهيز الخام والتحميص واختزال الخام في الفرن العالي وفرن الغاز وإنتاج الحديد الزهر والصلب وأنواع السبائك الثلاثة (وصف مفاهيمي للعمليات الصناعية) · 3 دروس
1

Iron ores and ore dressing

Chemistry Sec 3 Term 1, Chapter 1: pp. 16–20
GoalName the four iron ores with their formulas, compare their iron content by calculation, and describe ore dressing and roasting.
New words
ore · a rock from which a metal can be obtained at a profitore dressing · preparing the ore before reduction: crushing, sintering and concentratingroasting · heating the ore in air, which dries it, changes carbonate and hydrated compounds into oxide and burns sulphur and phosphorus away
Explanation

Iron ranks fourth among the elements of the Earth’s crust, with about 5 % by mass, behind oxygen, silicon and aluminium. Free iron is found mainly in meteorites, so almost all iron is taken from ores. An ore is chosen for its percentage of iron, for its impurities and for harmful elements in it, mainly sulphur, phosphorus and arsenic. The four ores are hematite Fe₂O₃, limonite 2Fe₂O₃·3H₂O, magnetite Fe₃O₄ and siderite FeCO₃. Ore dressing prepares them. Crushing breaks the ore into small pieces. Sintering joins fine ore into larger lumps. Concentrating removes waste rock; a magnetic ore such as magnetite can be separated with magnets. Roasting then heats the ore in air: it dries it, turns limonite and siderite into Fe₂O₃ and burns sulphur and phosphorus away as oxides. The equations are 2Fe₂O₃·3H₂O → 2Fe₂O₃ + 3H₂O, FeCO₃ → FeO + CO₂ then 4FeO + O₂ → 2Fe₂O₃, S + O₂ → SO₂ and 4P + 5O₂ → 2P₂O₅. They describe plant processes and are worked on paper only.

OreFormulaIron, % by mass (calculated)Note
hematiteFe₂O₃70.0red-brown
limonite2Fe₂O₃·3H₂O59.9yellow-brown, holds water
magnetiteFe₃O₄72.4black, magnetic
sideriteFeCO₃48.3a carbonate
Worked examples
ƒFormula
% Fe = (mass of Fe in the formula ÷ M) × 100
M = molar mass of the compound
Iron percentage of an ore compound. For Fe₂O₃ (Fe = 56, O = 16): M = 2 × 56 + 3 × 16 = 160 g/mol. The iron in one mole is 2 × 56 = 112 g, so the percentage is 112 ÷ 160 × 100 = 70.0 %.
Mass of iron in a sample. Two tonnes of pure hematite hold 2 × 0.700 = 1.4 tonnes of iron. If the ore were only 60 % Fe₂O₃, the iron would be 2 × 0.60 × 0.700 = 0.84 tonne, which is why ores are compared by their real iron content.
ƒEquation
FeCO₃(s) → FeO(s) + CO₂(g)
Roasting siderite on paper: FeCO₃ → FeO + CO₂ (Fe = 56, C = 12, O = 16). M(FeCO₃) = 116 g/mol, M(FeO) = 72 g/mol, M(CO₂) = 44 g/mol. So 116 g gives 72 g of FeO and 44 g of CO₂, and 72 + 44 = 116, so the mass is conserved.
Class activity

Three models of ore dressing, with a grown-up and no heating. Concentrating: mix paper clips into a bowl of lentils or rice and pick the clips out with a magnet inside a plastic bag; this models separating a magnetic ore. Crushing: break a dry cracker into crumbs. Sintering: press sugar crumbs into a lump with a few drops of water; this only models joining fine particles into a lump and is not how the factory does it. Write which step each model shows.

Practice
1
Calculate the iron percentage in FeO (Fe = 56, O = 16).
2
How much iron is in 5 tonnes of magnetite, if it is pure Fe₃O₄ (72.4 % Fe)?
3
Limonite roasts as 2Fe₂O₃·3H₂O → 2Fe₂O₃ + 3H₂O (M = 374, 160 and 18 g/mol). What mass of Fe₂O₃ and of water comes from 374 g of limonite?
4
Why must sulphur and phosphorus be removed from the ore before iron is made?