You have the stoichiometric equation. This tells you unequivocally that an
18
⋅
g
mass of water, 1 mole, reacts with a
56.07
⋅
g
mass of quicklime to form a
74.09
⋅
g
mass of slaked lime.
If you don't from where I am getting these numbers, you should know, and someone will be willing to elaborate.
Here, you have formed
6.21
⋅
m
o
l
of quicklime which requires stoichiometric lime AND water. And thus you need a mass of
6.21
⋅
m
o
l
×
18.01
⋅
g
⋅
m
o
l
−
1
water
≅
88
⋅
g
.
In practice, of course I would not weigh out this mass. I would just pour
100
−
200
⋅
m
L
of water into the lime.
