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ICSE Class 10 Metallurgy — Extraction of Metals Step by Step

By Aniket Bhardwaj · 21 September 2026 · CBSE / ICSE

Metallurgy is one of the chapters where ICSE Class 10 students lose marks not because the chemistry is hard, but because the vocabulary is slippery. Ore and mineral sound the same. Calcination and roasting sound the same. Flux and slag get swapped. This guide fixes the vocabulary first, then walks through the four stages of extraction, then does the aluminium process in full — because that is the one every ICSE paper keeps coming back to.

The five words you must define exactly

WordExact meaning
MineralAny naturally occurring compound (or native form) of a metal, whatever its purity.
OreA mineral from which the metal can be extracted profitably and conveniently. Every ore is a mineral; not every mineral is an ore.
Gangue (matrix)The earthy, unwanted rock mixed with the ore — sand, clay, mud.
FluxA substance added to combine with the gangue and make it fusible (easy to melt).
SlagThe fusible product formed when flux reacts with gangue. It floats on the molten metal and is drained off.
Flux + Gangue → Slag

If the gangue is acidic (silica, SiO2) you add a basic flux (limestone, CaO). If the gangue is basic (FeO) you add an acidic flux (silica). Example from the blast furnace: CaO + SiO2 → CaSiO3, calcium silicate slag.

Common ores worth memorising

MetalOreFormulaType
AluminiumBauxiteAl2O3·2H2OHydrated oxide
IronHaematiteFe2O3Oxide
ZincZinc blendeZnSSulphide
ZincCalamineZnCO3Carbonate
CopperCopper pyritesCuFeS2Sulphide
LeadGalenaPbSSulphide

Stage 1 — Concentration (dressing) of the ore

Removing gangue. Which method is used depends on a physical difference between the ore and the gangue:

Stage 2 — Converting the concentrated ore to the oxide

Metal oxides are easy to reduce, so carbonates and sulphides are converted to oxides first. This is where students lose the easiest marks in the chapter.

CalcinationRoasting
Air supplyLimited air / absence of airExcess of air
Ore typeCarbonate and hydrated oresSulphide ores
Gas given offCO2 and/or water vapourSO2
ExampleZnCO3 → ZnO + CO22ZnS + 3O2 → 2ZnO + 2SO2

Both equations above are balanced — count the atoms on each side and check for yourself. In the roasting equation: left side 2 Zn, 2 S, 6 O; right side 2 Zn, 2 O plus 2 S, 4 O — the same 2 Zn, 2 S, 6 O.

Stage 3 — Reduction of the oxide to the metal

The method is decided by the metal's position in the activity series. This is the single most examined idea in the chapter.

Position in activity seriesMetalsMethod of reduction
Most reactive (top)K, Na, Ca, Mg, AlElectrolysis of the molten compound — carbon is not a strong enough reducing agent
Moderately reactive (middle)Zn, Fe, Pb, CuReduction of the oxide by carbon or carbon monoxide
Least reactive (bottom)Ag, Au, HgOften found native; sulphide ores can be reduced by heat alone
ZnO + C → Zn + CO
Fe2O3 + 3CO → 2Fe + 3CO2

Stage 4 — Refining the metal

The commonest ICSE answer is electrolytic refining. For copper: the impure metal is the anode, a thin sheet of pure metal is the cathode, and the electrolyte is copper sulphate solution acidified with sulphuric acid.

Anode: Cu − 2e → Cu2+ (impure copper dissolves)
Cathode: Cu2+ + 2e → Cu (pure copper deposits)

Insoluble impurities such as silver and gold collect below the anode as anode mud.

Extraction of aluminium — the full ICSE answer

Step A — purification of bauxite (Bayer's process; printed as Baeyer's process in some ICSE books). Powdered bauxite is treated with hot concentrated sodium hydroxide. Alumina is amphoteric and dissolves; the iron(III) oxide impurity does not, and is filtered off as red mud.

Al2O3 + 2NaOH → 2NaAlO2 + H2O
NaAlO2 + 2H2O → NaOH + Al(OH)3↓  (on dilution and seeding)
2Al(OH)3 —heat→ Al2O3 + 3H2O

(Some syllabuses also list Hall's process and Serpeck's process as alternative purifications — check which ones your own textbook prints.)

Step B — electrolytic reduction (Hall–Héroult process). Pure alumina melts at a very high temperature, so it is dissolved in molten cryolite, Na3AlF6, with a little fluorspar, CaF2. The two additives lower the fusion temperature to roughly 950 °C and make the melt a much better conductor. The tank is lined with graphite (the cathode) and graphite rods dip in as anodes; a low voltage with a very large current is used.

Cathode: Al3+ + 3e → Al
Anode: 2O2− − 4e → O2
Overall: 2Al2O3 → 4Al + 3O2

The oxygen attacks the hot graphite anodes (C + O2 → CO2), so they burn away and have to be replaced from time to time. That is a favourite one-mark question.

Worked example 1 — how much aluminium is in the alumina?

Q. Calculate the percentage of aluminium in pure alumina, Al2O3, and hence the mass of aluminium obtainable from 1000 kg of it. (Al = 26.982, O = 15.999)

Molar mass:
Al: 2 × 26.982 = 53.964
O: 3 × 15.999 = 47.997
M(Al2O3) = 53.964 + 47.997 = 101.961 g/mol

% Al = (53.964 ÷ 101.961) × 100 = 0.52927 × 100 = 52.93 %

Mass of Al from 1000 kg = 1000 × 0.5293 = 529.3 kg

Check: % O = (47.997 ÷ 101.961) × 100 = 47.07 %, and 52.93 + 47.07 = 100.00 ✓

Worked example 2 — zinc from zinc blende

Q. 500 g of pure zinc blende is roasted and the zinc oxide formed is completely reduced by carbon. What mass of zinc is obtained? (Zn = 65.38, S = 32.06)

The two equations are
2ZnS + 3O2 → 2ZnO + 2SO2
ZnO + C → Zn + CO
so 1 mol ZnS finally gives 1 mol Zn.

M(ZnS) = 65.38 + 32.06 = 97.44 g/mol
Moles of ZnS = 500 ÷ 97.44 = 5.1314 mol
Moles of Zn = 5.1314 mol
Mass of Zn = 5.1314 × 65.38 = 335.5 g

Second route as a check: mass ratio Zn : ZnS = 65.38 ÷ 97.44 = 0.6710, and 0.6710 × 500 = 335.5 g. The two routes agree.

Worked example 3 — choosing the method

Q. Name the method used to extract (a) sodium from molten sodium chloride, (b) iron from haematite, (c) copper from copper pyrites after roasting. Give a reason in each case.

(a) Electrolysis of the molten chloride. Sodium is near the top of the activity series, so its compounds are very stable and carbon cannot reduce them.

(b) Reduction of Fe2O3 by carbon monoxide in the blast furnace. Iron is in the middle of the series, so a carbon-based reducing agent is strong enough and much cheaper than electricity.

(c) Roasting followed by self-reduction, then electrolytic refining. Copper is low in the series, so its oxide is easy to reduce; refining is electrolytic because copper for wiring must be very pure.

Alloys — the short answers ICSE asks for

An alloy is a homogeneous mixture of a metal with one or more other elements. Alloying usually increases hardness and resistance to corrosion, and lowers the melting point.

AlloyMetals presentTypical use
DuraluminAluminium, copper, magnesium, manganeseAircraft and vehicle parts
MagnaliumAluminium, magnesiumLight instrument frames
BrassCopper, zincFittings, utensils
BronzeCopper, tinStatues, coins, medals

Percentage compositions differ from book to book, so quote the figures printed in your own textbook rather than a number you half-remember.

Mistakes that cost marks

  • Calling every mineral an ore. Ore = the mineral we actually extract the metal from, profitably. Write the "profitably and conveniently" part — it is the mark.
  • Swapping calcination and roasting. Remember: calcination for carbonates (limited air); roasting for sulphides (excess air, SO2 out).
  • Using froth flotation for an oxide ore. It works only because sulphide surfaces are preferentially wetted by oil. Bauxite is never concentrated this way.
  • Saying cryolite is the ore of aluminium in the Hall–Héroult cell. In that cell cryolite is the solvent. The aluminium comes from the alumina.
  • Forgetting why graphite anodes are replaced. They are burnt away by the oxygen liberated at them, not "worn out by the current".
  • Mixing up flux and slag. Flux is what you add; slag is what comes out.

Where this chapter is examined

Board / examHow it appears
ICSE Class 10A named chapter: definitions, the four stages, the aluminium process with electrode equations, alloys
CBSE Class 10Inside "Metals and Non-metals" — activity series decides the extraction method, plus roasting/calcination
Class 12The same ideas reappear with thermodynamics attached (Ellingham diagrams)
Competitive papersOre-to-metal matching and yield calculations of exactly the kind worked above

Qualitatively, metallurgy is one of the highest-return chapters in ICSE chemistry because so much of it is recall rather than reasoning — but do not treat the numericals as optional. Yield questions like the two worked above are ordinary mole arithmetic in disguise.

Check the metal content of any ore instantly. Type a formula such as Al2O3, Fe2O3, ZnS or CuFeS2 into the Molar Mass & Composition tool and it returns the molar mass and the mass percentage of every element — exactly the number Worked Example 1 calculates by hand.

Open the Molar Mass & Composition Calculator →

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