39.6k views
1 vote
To neutralized 1.65g LiOH, how much .150M HCl would be needed?

1 Answer

5 votes

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.


User Smit Bhanvadia
by
8.0k points

Related questions

1 answer
4 votes
208k views
asked Jul 4, 2019 206k views
N Raghu asked Jul 4, 2019
by N Raghu
8.0k points
2 answers
2 votes
206k views
asked Jan 13, 2019 63.3k views
Shree asked Jan 13, 2019
by Shree
7.9k points
1 answer
1 vote
63.3k views
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.