Question

Aluminum metal will reduce water, but the reaction is extremely slow under neutral conditions at room temperature.

                Al(s) + 3H₂O(l) → Al(OH)₃(s) + H₂(g)                             very, very slow

In the presence of aqueous base, however, Al will readily react to form the very soluble Al(OH)₄^–1(aq) complex anion which can be neutralized with H₂SO₄(aq) in two steps. The first produces the very insoluble Al(OH)₃(s). If the Al(OH)₃(s) slurry is then immediately boiled with additional H₂SO₄(aq) the insoluble hydroxide will react completely to give a clear, colorless solution. After excess water is boiled off, cooling produces KAl(SO₄)₂^.12H₂O(s), Alum, as fine, white needles.   

                Al(s)    +     3H₂O     +   KOH           ⟶    KAl(OH)₄(aq) + H₂(g)   fast and exothermic at room temperature

                KAl(OH)₄(aq)   +   ½ H₂SO₄(aq)     ⟶    Al(OH)₃(s) + H₂O(l) + ½ K₂SO₄(aq)

                Al(OH)₃(s)     +     H₂SO₄(aq)      ⟶   ½ Al₂(SO₄)₃(aq) + 3 H₂O   (requires heating)

                ½ Al₂(SO₄)₃(aq)   +   ½ K₂SO₄(aq) ⟶ KAl(SO₄)₂(s)

Combining these four equations gives the overall reaction.

                Al(s)   + KOH(aq) + 2 H₂SO₄(aq) ⟶   [KAl(SO₄)₂(aq)] + H₂O(l) + H₂(g) and

                [KAl(SO₄)₂(aq)] + 12 H₂O(l) ⟶   KAl(SO₄)₂^.12H₂O(s) (Alum)

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure. Calculate the volume of 6.00 M H2SO4 (aq) needed just to precipitate Al(OH)3 (s).

Answer 1

   Al(s) + 3H2O + KOH ---> KAl(OH)4(aq) + H2(g)

   fast and exothermic at room temperature

KAl(OH)4(aq)   +   ½ H2SO4(aq)   -->    Al(OH)3(s) + H2O(l) + ½ K2SO4(aq)

from equation :1 mol Al = 1 mol KOH = 1/2 mol H2SO4

no of mol of Al reacted = w/mwt = 0.946/27 = 0.035 mol

no of mol of KOH reacted = 35*2/1000 = 0.07 mol

limiting reactant = Al

no of mol of H2SO4 required = 0.035*1/2 = 0.0175 mol

volume of H2SO4 required = n/M = 0.0175/6 = 0.00292 L

                          = 2.92 ml