Respuesta :
Answer: 0.458 liters of HCl will be needed to neutralize LiOH.
Explanation: Reaction follows:
[tex]HCl(aq.)+LiOH(aq.)\rightarrow LiCl(s)+H_2O(l)[/tex]
Stoichiometry ratio of LiOH and HCl is 1 : 1
which means, 1 mole of LiOH reacts with 1 mole of HCl.
Number of moles can be calculated by
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
mass of LiOH = 1.65g (Given)
molar mass of LiOH = 24g
[tex]\text{Number of moles}=\frac{1.65g}{24g}[/tex]
Number of moles = 0.06875 moles
Number of moles of LiOH = Number of moles of HCl
Moles of HCl = 0.06875 moles
and [tex]Molarity=\frac{Moles}{Volume}[/tex]
Molarity = 0.15 moles/Liter (Given)
Volume needed to neutralize LiOH will be,
[tex]Volume=\frac{Moles}{Molarity}[/tex]
[tex]V=\frac{0.06875moles}{0.15moles/litre}[/tex]
Volume of HCl needed to neutralize LiOH = 0.458 litre.
