Ask Question
10 February, 02:38

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

+5
Answers (1)
  1. 10 February, 02:49
    0
    The molecular weight of LiOH would be 23.95 g/mol, so the amount of LiOH in mol would be: 1.64g / (23.95 g/mol) = 0.069 mol

    The reaction of LiOH with HCl would be:

    HCl + LiOH = H2O + LiCl

    The coefficient of LiOH:HCL is 1:1 so you need the same amount of HCl to neutralize LiOH.

    HCl = LiOH

    volume * 0.15M = 0.069 mol

    volume = 0.069 mol / (0.15 mol / 1000ml)

    volume = 459.29 ml
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “To neutralize 1.65g LiOH, how much. 150 M HCl would be needed ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers