help with answering this limiting reactant question without converting between grams and moles?

Question

Equation CaO+H2O-->Ca(OH)2
Question: how much calcium hydroxide will be produced when 10g of calcium oxide reacts with 3g of water?

I belive rather than having to covert from grams to moles there is an alternate method:
CaO+H2O-->Ca(OH)2
56.08g : 18.01 g
10.00g : Xg now I am unsure what to do from here!
12.34g is the final answer
Any help on this alternate method is appreciated

Answer 1

An alternate method is not an alternate method unless it produces a result. What you set out to do is OK, and what you are solving for a X is the amount of water that would completely react with 10 g of CaO. If this is more than 3 grams, you have to reverse course and take the proportion as
56.08/X = 18.01/3, and X is the amount of CaO that is reacted with 3 grams of water. To find Ca(OH)2, you would have to make yet another proportion:
Mole wt Ca(OH)2/56.08 * X = answer.

Actually, your alternate is really the moles in disguise, since the reference masses are molar masses and you follow the chem rxn.

Answer 2

CaO + H2O ----------> Ca(OH)2

Moles CaO = 10 g / 56 g/mole = 0.179 mole

Moles H2O = 3 g / 18 g/mole = 0.167 mole

Since CaO and H2O react in a 1 :1 mole ratio, there is more CaO than H2O. So water is the limiting reactant. And since there is a 1 :1 mole ratio between water and Ca(OH)2, you can only form 0.167 moles of calcium hydroxide.

Mass Ca(OH)2 formed = 0.167 mole X 74 grams / mole = 12.33 grams