Question

What volume of 0.100 M Ca(OH)2 is required to neutralize 100.0 mL of a 0.200 M HCI solution?

100mL?

Answer 1

I have an answer below the question I just needed to know if it was right or not..

Answer 2

yes it should be correct

Answer 3

Starting from following korosh23's steps, we shall have:
Ca(OH)2 + 2HCl - CaCl2 + 2H2O.
Aditionally, when they say ' is needed to react with ...' It means the substance that is to be mentioned is the limiting reagent.
For such substace, find the moles, using the formula n=cv/1000, where n is moles, c is concentration and v is volume in cm^3.
Hence, we shall have n=0.2*100/1000= 0.02mol.
According to our equation, the ratio of Ca(OH)2 to HCl is 1:2, therefore we double the moles found to be one available for the reaction which is 2*0.02= 0.04 mol of Ca(OH)2. Using this moles, we reuse the formula of mol to find v.
V=1000*n/c= 1000*0.04/0.1= 400mL.

Answer 4

The equation will be Ca(OH)2 + 2HCl - CaCl2 + 2H2O
The molar ratio is 1:2
To find the moles of the acid, we use the formula n=cv/1000, where n is mol, c is concentration and v is volume in cm^3.
n= 0.2*100/1000=0.02mol of HCl
Then the mol of the base will be 0.02*1/2= 0.01 mol
Using the mol formula , making v the subject will be v=1000*n/c= 1000*0.01/0.1

This is what I had

Answer 5

Basically I did exactly what you did except I divided

Answer 6

Oh Yes. My mistake! I doubled instead of halving it. You got it right, @ZDoubleM.

Answer 7

Haha awesome, thanks mate