Aluminium — chemistry, alloys; alumina, alums: Worked examples

This page drills the parent topic the way an exam actually attacks it: not one clean question, but every awkward corner of aluminium chemistry. First we map the corners; then we walk each one.

Before we start, one rule we will use again and again:

The scenario matrix

Every question the p-Block chapter throws at aluminium falls into one of these cells. Each worked example below is labelled with the cell(s) it clears.

#	Case class	The "sign / quadrant" here	Cleared by
C1	Reagent = acid (dilute)	acid side of amphoterism	Ex 1
C2	Reagent = base (alkali)	base side of amphoterism	Ex 2
C3	Degenerate / "nothing happens" input (conc. $H N O_{3}$ )	zero-reaction case	Ex 3
C4	Redox as reducing agent (thermite)	thermodynamic sign of $Δ H$	Ex 4
C5	Structure / electron-counting limit ( $A l_{2} C l_{6}$ )	electron-deficient extreme	Ex 5
C6	Aqueous hydrolysis (pH sign)	acidic-solution corner	Ex 6
C7	Formula / stoichiometry of an alum	different-element twist	Ex 7
C8	Real-world word problem (water purification mass)	applied quantitative	Ex 8
C9	Exam twist (spot the contradiction)	trap / "both wrong" case	Ex 9

Ex 1 — The acid side (cell C1)

Forecast: guess first — will it be more or less than $22.4 L$ ? Each Al gives $1.5$ of an $H_{2}$ , and we have only $0.1 mol$ Al, so expect a few litres, not a full mole's worth.

Write the balanced reaction. $2 A l + 6 H C l \to 2 A l C l_{3} + 3 H_{2} ↑$ . Why this step? Because passivation does not save Al from dilute acid — the acid strips fresh metal, so this is the ordinary metal-plus-acid reaction, not a trick case.
Moles of Al $= \frac{2.70}{27} = 0.100 mol$ . Why this step? Grams tell us nothing until converted to moles — moles are the "counting unit" the equation is written in.
Mole ratio $A l : H_{2} = 2 : 3$ , so $n_{H_{2}} = 0.100 \times \frac{3}{2} = 0.150 mol$ . Why this step? The coefficients are a recipe; $2$ Al always makes $3 H_{2}$ .
Volume $= 0.150 \times 22.4 = 3.36 L$ . Why this step? At STP one mole of any gas occupies $22.4 L$ .

Verify: Units: $mol \times L mol^{- 1} = L$ ✓. And $3.36 L < 22.4 L$ , matching our forecast. ✓

Ex 2 — The base side, the amphoteric proof (cell C2)

Forecast: Al is amphoteric — it fights base too. Guess: the same volume, because the electrons Al loses (3 per atom) don't care whether the other side is acid or base.

Balanced reaction: $2 A l + 2 N a O H + 6 H_{2} O \to 2 N a [A l (O H)_{4}] + 3 H_{2} ↑$ . Why this step? $O H^{-}$ dissolves the oxide skin as the aluminate ion $[A l (O H)_{4}]^{-}$ (a coordination species), exposing fresh metal — the signature of amphoterism.
Same ratio $A l : H_{2} = 2 : 3$ , same $0.100 mol$ Al. Why this step? Al is still oxidised $0 \to + 3$ , losing $3 e^{-}$ ; the $H_{2}$ count is fixed by that electron balance.
$n_{H_{2}} = 0.150 mol \Rightarrow V = 3.36 L$ .

Verify: Identical to Ex 1 ✓ — this is the punchline: acid and base give the same gas, proving Al is amphoteric.

Ex 3 — The degenerate "nothing happens" case (cell C3)

Forecast: A strong oxidiser — surely violent? Careful: this is the trap cell.

Recognise passivation. Conc. $H N O_{3}$ instantly grows a dense $A l_{2} O_{3}$ armour. Why this step? The parent [!mistake] warns us: the oxidiser's first act is to build the impervious oxide film — see Passivation and Corrosion.
Consequence: the film seals the metal; reaction stops. Why this step? No fresh Al surface $\Rightarrow$ no further attack.
$n_{H_{2}} = 0$ , so $V = 0 L$ .

Verify: This is the "zero input" corner of our matrix. Sanity check: conc. $H N O_{3}$ is literally transported in aluminium tankers — consistent with $V = 0$ . ✓

Ex 4 — Reducing agent, the thermite sign check (cell C4)

Forecast: Al binds oxygen more hungrily than Fe does (see Ellingham), so expect a large negative $Δ H$ (heat released, molten iron).

Hess's law: $Δ H_{r x n} = [Δ H_{f} (A l_{2} O_{3}) + 2Δ H_{f} (F e)] - [2Δ H_{f} (A l) + Δ H_{f} (F e_{2} O_{3})]$ . Why this step? Elements ( $A l$ , $F e$ ) have $Δ H_{f} = 0$ , so only the two oxides survive.
Substitute: $Δ H_{r x n} = (- 1676) - (- 824) = - 852 kJ$ . Why this step? This is the swap of oxygen from the weaker holder (Fe) to the stronger holder (Al).
Sign: $Δ H_{r x n} < 0 \Rightarrow$ exothermic $\Rightarrow$ Al is the reducing agent, the oxide goes to the more stable $A l_{2} O_{3}$ .

Verify: $- 1676 + 824 = - 852 kJ$ ✓, large and negative — matches "molten iron for welding rails." ✓

Ex 5 — Electron-counting limit: the $A l_{2} C l_{6}$ dimer (cell C5)

Forecast: Al in $A l C l_{3}$ is short of a full octet — guess it will "borrow" a lone pair.

Figure — Aluminium — chemistry, alloys; alumina, alums

Count around monomeric Al. Three $A l - C l$ bonds $= 3 \times 2 = 6$ electrons. Octet needs $8$ : short by 2 — one empty orbital (red gap in the figure). Why this step? An empty orbital on Al makes it a Lewis acid (electron-pair acceptor).
Donate a lone pair. A terminal Cl on a second molecule pushes one of its lone pairs (cyan arrow) into that empty orbital — a coordinate (dative) bond. Why this step? Chlorine has three lone pairs to spare; giving one completes Al's octet.
Result: two such bridges form, giving $A l_{2} C l_{6}$ : 2 bridging Cl, 4 terminal Cl, each Al now $s p^{3}$ with $8$ electrons. Each bridging Cl forms 2 Al–Cl bonds (one normal, one dative).

Verify: Total Cl $= 2 bridge + 4 terminal = 6 = 2 \times 3$ ✓ — mass-balances two $A l C l_{3}$ units. Each Al: $4$ bonds $\times 2 e = 8$ ✓ octet complete.

Ex 6 — Aqueous hydrolysis: the pH sign (cell C6)

Forecast: $C l^{-}$ is the conjugate base of a strong acid (inert). The action is all on $A l^{3 +}$ — guess acidic.

Al³⁺ is hydrated: $[A l (H_{2} O)_{6}]^{3 +}$ . Why this step? The high charge density (our master key) pulls six water molecules in tight.
Polarise the O–H bonds. The $+ 3$ centre drags electron density off the coordinated water, loosening an $O - H$ bond: $[A l (H_{2} O)_{6}]^{3 +} ⇌ [A l (H_{2} O)_{5} (O H)]^{2 +} + H^{+}$ Why this step? Same polarising power that makes $A l C l_{3}$ covalent now squeezes protons out of water.
Verdict: free $H^{+}$ released $\Rightarrow$ acidic, $p H < 7$ . Overall the parent writes it compactly as $A l C l_{3} + 3 H_{2} O \to A l (O H)_{3} + 3 H C l$ .

Verify: Consistency check — a small, highly charged cation → acidic salt; a large low-charge cation (e.g. $N a^{+}$ ) → neutral salt. $A l^{3 +}$ sits firmly in the acidic corner. ✓

Ex 7 — Formula twist: is $K C r (S O_{4})_{2} \cdot 12 H_{2} O$ an alum? (cell C7)

Forecast: The parent [!mistake] warns the two cations only need different charges, not the same element. Guess: yes, it is an alum.

Recall the template: $M^{+} M^{' 3 +} (S O_{4})_{2} \cdot 12 H_{2} O$ (equivalently $M_{2} S O_{4} \cdot M_{2}^{'} (S O_{4})_{3} \cdot 24 H_{2} O$ ). Why this step? An alum is defined by its charge pattern, not a particular metal.
Match charges. $K$ is $+ 1$ ( $M^{+}$ ); $C r$ here is $+ 3$ ( $M^{' 3 +}$ ); anion $(S O_{4})_{2}$ is $2 \times (- 2) = - 4$ . Why this step? Charge must balance overall.
Balance check: $(+ 1) + (+ 3) + (- 4) = 0$ ✓ — neutral. So chrome alum is a valid alum.

Verify: Charge sum $= 0$ ✓. Contrast potash alum $K A l (S O_{4})_{2} \cdot 12 H_{2} O$ : $(+ 1) + (+ 3) - 4 = 0$ ✓ — same pattern, different trivalent metal. The student's claim is false.

Ex 8 — Real-world word problem: dosing a water tank (cell C8)

Forecast: Each formula unit gives exactly one $A l^{3 +}$ , so mass $\approx 0.02 \times 474 \approx 9.5 g$ .

Confirm molar mass: $39 + 27 + 2 (32 + 64) + 12 (18) = 39 + 27 + 192 + 216 = 474 g mol^{- 1}$ . Why this step? We must trust the $M$ we plug in; count every atom including the $12$ waters.
One Al per unit: $n_{alum} = n_{A l^{3 +}} = 0.020 mol$ . Why this step? The subscript on $A l$ is $1$ ; moles of salt = moles of $A l^{3 +}$ .
Mass $= 0.020 \times 474 = 9.48 g$ . Why this step? $mol \times g mol^{- 1} = g$ .

Verify: $0.020 \times 474 = 9.48 g$ ✓, matching the forecast. The dissolved $A l^{3 +}$ then forms gelatinous $A l (O H)_{3}$ that traps dirt and settles — coagulation. ✓

Ex 9 — Exam twist: spot the false statement (cell C9)

Forecast: Three of these are core facts we've just proved; one is the classic trap.

Test (a): $A l_{2} O_{3} + 6 H C l \to 2 A l C l_{3} + 3 H_{2} O$ and $A l_{2} O_{3} + 2 N a O H + 3 H_{2} O \to 2 N a [A l (O H)_{4}]$ . True (amphoteric).
Test (b): conc. $H N O_{3}$ passivates Al (Ex 3) → no $H_{2}$ . FALSE. ✗ Why this step? This is exactly the C3 degenerate case; "rapid $H_{2}$ " contradicts passivation.
Test (c): Ex 5 established 2 bridging + 4 terminal Cl. True.
Test (d): Ex 6 established acidic hydrolysis. True.

Verify: Only statement (b) contradicts an established result — it is the wrong one. ✓ (It even claims $H_{2}$ where Ex 3 gave $0 L$ .)

Recall Self-test before you close the page

Volume of $H_{2}$ from $2.70 g$ Al in dilute $H C l$ at STP? ::: $3.36 L$ Same Al in $N a O H$ — more, less, or equal? ::: Equal, $3.36 L$ (amphoterism) Same Al in conc. $H N O_{3}$ ? ::: $0 L$ — passivated $Δ H$ of the thermite reaction (from Ex 4)? ::: $- 852 kJ$ , exothermic Grams of potash alum for $0.020 mol$ $A l^{3 +}$ ? ::: $9.48 g$ Is $K C r (S O_{4})_{2} \cdot 12 H_{2} O$ an alum? ::: Yes — cations are $+ 1$ and $+ 3$

Connections

3.2.02 Aluminium — chemistry, alloys; alumina, alums (Hinglish) — the parent topic these examples drill
Fajans Rules — the charge-density master key behind Ex 5, 6
Lewis Acids and Bases — electron-deficient $A l C l_{3}$ (Ex 5)
Passivation and Corrosion — the C3 zero-reaction case (Ex 3, 9)
Thermodynamics of Reduction (Ellingham) — why $Δ H < 0$ in Ex 4
Coordination Compounds — the aluminate ion in Ex 2
Group 13 Elements — Al as reducing agent