Hello!
The chemical reaction for the neutralization of LiOH is the following:
LiOH + HCl → LiCl + H₂O
To calculate how much 0,150 M HCl would be needed, we can apply the following conversion factor, to go from grams of LiOH to mL of HCl
[tex]1,65 g LiOH* \frac{1 mol LiOH}{23,95 g LiOH}* \frac{1 mol HCl}{1 mol LiOH} * \frac{1 L Sol}{0,150 mol HCl}* \frac{1000 mL}{1L} \\ \\ =459,29 mL HCl [/tex]
So, 459,29 mL of 0,150 M HCl are required to neutralize 1,65 g of LiOH
Have a nice day!
