Question

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

Answer 1

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