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

Answer:

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.