Question

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

Answer 1

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

Answer 2

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