To neutralize 1.65g LiOH, how much. 150 M HCl would be needed

The molecular weight of LiOH would be 23.95 g/mol, so the amount of LiOH in mol would be: 1.64g / (23.95 g/mol) = 0.069 mol

The reaction of LiOH with HCl would be:

HCl + LiOH = H2O + LiCl

The coefficient of LiOH:HCL is 1:1 so you need the same amount of HCl to neutralize LiOH.

HCl = LiOH

volume * 0.15M = 0.069 mol

volume = 0.069 mol / (0.15 mol / 1000ml)

volume = 459.29 ml