help with solving a limiting reagent question using two different methods?

Question

I have to solve the following limiting reagent question using two different methods and am stuck! I think I can use gram to mole conversation and dimensional analysis? I am new to this stuff so any help most appreciated.
CaO + H2O -> Ca(OH)2
find out how much calcium hydroxide will be produced when 10.00 g of calcium oxide reacts with 3.00 g of water.

Answer 1

I think that the first answerer used a correct method but made a mistake. Therefore I will try to show you two methods.

The molar mass of CaO is 56.077g/mol
The molar mass of H2O is 18.015g/mol
The molar mass of Ca(OH)2 is 74.092g/mol

Method 1:
Let us figure out how many moles of each reactant we have to start with:
10.00g/56.077g/mol = 0.1783 mol CaO
3.00g/18.015g/mol = 0.1665 mol H2O
There is less water than CaO, so water is the limiting reactant. Thus the maximum amout of Ca(OH)2 that can be formed is 0.167 mol:
0.1665mol*74.092g/mol = 12.3g Ca(OH)2

Method 2:
We will use simple ratio to see how many grams of Ca(OH)2 can be formed if either 10.00g CaO or 3.00g H2O is fully reacted. We know from the molar masses that 56.08g of CaO and/or 18.02g of H2O fully reacted would form 74.09g Ca(OH)2:
CaO ..+.. H2O ==> Ca(OH)2
56.08g...18.02g......74.09g
10.00g....................X(g)
..............3.00g........Y(g)
X/10.00 = 74.09/56.08, or X = 74.09*10.00/56.08 = 13.2
Y/3.00 = 74.09/18.02, or Y = 74.09*3.00/18.02 = 12.3
Since Y < X, water is the limiting reactant, and Y = 12.3(g) (of Ca(OH)2) is the answer.

Answer 2

Don't know about two methods but can help you get the answer:)

MW of CaO is 56.077g/mol
MW of H2O is 18.0148 g/mol
MW Ca(OH)2 is 58.0928 g/mol

Let us how many moles of each reactant we have to start with:

10.00g/56.077g/mol = 0.1783 mol CaO
3.00g/18.0148g/mol = 0.167 mol H2O

there is less water than CaO (mole-wise), so water is the limiting reactant. Thus the maximum amout of Ca(OH)2 that can be formed is 0.167 mol.

let us convert that into grams: 0.167mol*58.0928g/mol = 9.70g Ca(OH)2

hope that was helpful