将0.1molAl投入3mol/LNaOH溶液100mL中，

2Al + 2OH- + 6H2O ----> 2[Al(OH)4]- + 3H2

to completely react with 0.1 mol of Al, 0.1 mol OH- is required. There is still excess OH- left behind, of which the mol = 0.2 mol.
To neutralize 0.2 mol OH- requires 1 M H2SO4
V = 0.2/2 =0.1 L = 100 mL

Since the final precipitate is Al(OH)3
[Al(OH)4]- + H+ ----> Al(OH)3 + H2O
to completely convert [Al(OH)4]- to Al(OH)3 requires
V'=0.1/2 =0.05 L = 50 mL

so V(total) = V + V' = 150 ml of H2SO4

忘了 高中学的 先生成偏铝酸钠，然后再和酸反应，应该很简单的