To neutralize 1.65g lioh, how much .150 m hcl would be needed?

To  neutralize  1.65  LiOH  , how  the HCL  needed  is  calculated  as  follows

calculate  the  moles  of LiOH   been neutralized

 1.65/23.95 = 0.069  moles

write  the   reaction   equation
HCl  + LiOH =  LiCl +H2O

by   use  of  mole  ratio   between  HCL  to LiOH  that  is  1:1  the  the  moles of  HCl  is  therefore =  0.069 moles

The HC needed   is  therefore  =  moles/ molarity

=  0.069/0.150 = 0.46  liters

The reaction between LiOH and HCl is;
   LiOH + HCl → LiCl + H₂O

the stoichiometric ratio between LiOH and HCl is 1 : 1

Molar mass of LiOH = 24 g/mol

LIOH moles = mass / molar mass = 1.65 / 24 = 0.06875 mol

Hence needed HCl moles = 0.06875 mol

according to the given HCl concentration, 1 L volume has 0.150 mol of HCl.
Hence HCl volume = moles / concentration
                               = 0.06875 mol / 0.150 mol/L
                               = 0.458 L = 458 mL

How do we find the answer?

The answer for this question is d. none of these