The reaction between LiOH and HCl is;
LiOH + HCl → LiCl + H₂O
The stoichiometric ratio between LiOH and HCl is 1 : 1
moles of LiOH added = moles of HCl needed to neutralize.
Molar mass of LiOH = 24 g/mol
moles = mass / molar mass
LIOH moles = 1.65 / 24 = 0.06875 mol
Hence needed HCl moles = 0.06875 mol
Molarity = moles (mol) / Volume (L)
Hence needed HCl volume = moles / molarity
= 0.06875 mol / 0.150 mol/L
= 0.458 L = 458 mL
