Experiment4: Preparation of alum [KAl(SO4)2.12.H2O]
Stoichiometry and the limited reactant concept
The mechanism of formation of alum
2 Al(s) + 2 KOH(aq) + 6 H2O(l) → 2 K[Al(OH)4](aq) + 3 H2(g) (1)
2 K[Al(OH)4](aq) + H2SO4(aq) → 2 Al(OH)3(s) + 2 H2O (l)+ K2SO4(aq) (2)
2 Al(OH)3(s) + 3 H2SO4 → Al2(SO4)3(aq) + 6 H2O (3)
Al2(SO4)3 (aq) + K2SO4(aq) + 24H2O → 2 KAl(SO4)2•12H2O(s) (4)
2 Al(s) + 2 KOH(aq) + 4 H2SO4 + 22 H2O → 2 KAl(SO4)2•12H2O(s) + 3 H2(g) overall reaction
Stoichiometry of the overall reaction
2 moles Al(s) 2 moles alum
2 moles KOH(aq) 2 moles alum
4 moles H2SO4(aq) 2 moles alum
Calculations
1. Calculate number of moles of Al(s):
(g/mol) 98.26
m
)Al(MM
m
n(g) Al
Al
Al
2. Calculate number of moles of KOH:
KOH
KOH VMn
3. Calculate number of moles of H2SO4:
42
42 SOH
SOHVMn
4. Use stoichiometry of the balanced overall reaction to determine the limiting reactant and
the theoretical yield in moles and grams
5. Calculate the % yield