How many grams of water react to form 7.21 moles of ca(oh)2

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.

with this one i know the answer in craters one reason i looked it up and i learned about it there is my evidence