Question

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

Answer 1

Answer: 0.458 liters of HCl will be needed to neutralize LiOH.

Step-by-step explanation: Reaction follows:

`HCl(aq.)+LiOH(aq.)\rightarrow LiCl(s)+H_2O(l)`

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

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}`

mass of LiOH = 1.65g (Given)

molar mass of LiOH = 24g

`\text{Number of moles}=(1.65g)/(24g)`

Number of moles = 0.06875 moles

Number of moles of LiOH = Number of moles of HCl

Moles of HCl = 0.06875 moles

and
`Molarity=(Moles)/(Volume)`

Molarity = 0.15 moles/Liter (Given)

Volume needed to neutralize LiOH will be,

`Volume=(Moles)/(Molarity)`

`V=(0.06875moles)/(0.15moles/litre)`

Volume of HCl needed to neutralize LiOH = 0.458 litre.