# 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)?

##### 1 Answer
Dec 1, 2016

Clearly, I need $6.21$ equiv, $6.21 \cdot \text{moles}$ of water. Why clearly?

#### Explanation:

You have the stoichiometric equation. This tells you unequivocally that an $18 \cdot g$ mass of water, 1 mole, reacts with a $56.07 \cdot g$ mass of quicklime to form a $74.09 \cdot g$ 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 $6.21 \cdot m o l$ of quicklime which requires stoichiometric lime AND water. And thus you need a mass of $6.21 \cdot \cancel{m o l} \times 18.01 \cdot g \cdot \cancel{m o {l}^{-} 1}$ water $\cong 88 \cdot g$.

In practice, of course I would not weigh out this mass. I would just pour $100 - 200 \cdot m L$ of water into the lime.