Question

How many grams of water react to form 6.21 moles of `Ca(OH)_2` in `CaO(s) + H_2O(l) -> Ca(OH)_2(s)`?

Answer 1

Clearly, I need `6.21` equiv, `6.21* "moles"` of water. Why clearly?

Explanation:

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*mol` of quicklime which requires stoichiometric lime AND water. And thus you need a mass of `6.21*cancel(mol)xx18.01*g*cancel(mol^-1)` water `~=88*g`.

In practice, of course I would not weigh out this mass. I would just pour `100-200*mL` of water into the lime.