preparation of alum Using 8.0 x 10-2 mol KOH and 1.5 X 10-1 mol H2SO4.Reagent is in excess + amount?

Answer 1

The reaction is:
2 KOH + H2SO4 --> K2SO4 + H2O

You need 2 moles of KOH for every mole of H2SO4 reacted.
You have 0.08 mol of KOH and 0.15 mol of H2SO4, so the reagent in excess is H2SO4. I'm not really sure what you're asking, but:
the final concentrations will be:
KOH: 0 mol
H2SO4: 0.11 mol
K2SO4: 0.04 mol

You reacted 0.08 mol of KOH and 0.04 mol of H2SO4.

Hope that helped.

Answer 2

The preparation of alum is a 2-step process. The KOH is used in the first step:

2 Al (s) + 2 KOH (aq) + 6 H2O (l) --> 2 KAl(OH)4 (aq) + 3 H2 (g)

One mole of KOH is required for each Tetrahydroxo-aluminate complex formed. 0.08 moles of KOH are used.

The second step is a neutralization:

KAl(OH)4 (aq) + 2 H2SO4 (aq) + 8 H2O (l) --> KAl(SO4)2*12 H2O (s)

Two moles of Sulfuric acid are required for each Tetrahydroxo-aluminate complex that reacts. 0.16 moles would be needed, but only 0.15 moles of H2SO4 were used. This is the limiting reagent.

0.15 moles / 2 = 0.075 moles of KOH would be needed, The excess is:

0.08 - 0.075 = 0.005 moles KOH in excess.