Question

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

Answer 1

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


`1,65 g LiOH* (1 mol LiOH)/(23,95 g LiOH)* (1 mol HCl)/(1 mol LiOH) * (1 L Sol)/(0,150 mol HCl)* (1000 mL)/(1L) \\ \\ =459,29 mL HCl`

So, 459,29 mL of 0,150 M HCl are required to neutralize 1,65 g of LiOH

Have a nice day!