How come the balanced equation for calcium oxide reaction with water is: cao (s) +h2o (l) --> ca(oh)2 (s)?

CaO(s) + H2O(l) ---> Ca(OH)2(aq) 
CaO(s) + H2O(l) ---> 2CaOH(aq) we get this after balancing this equation
2CaO(s) + H2O(l) ---> 2CaOH(aq) 
 number of moles
CaO 

n = given mass / molar mass. 
Ca = 40 
O = 16 
Mass CaO = 56 grams/mole 

n = 2.50 / 56 grams/mole = 0.0446 moles 
Molar mass = 2*H + O = 2 + 16 = 18 grams/ mol 
now in water
n = 2.41 / 18 = 0.134 
moles of water - moles of water used in the reaction = left over moles. 

0.134 - 0.0446 = 0.0893 remaining number of moles

 grams of water which are left 

1 mole of water = 18 grams 
0.0893 moles of water = x 

x = 18 * 0.0893 = 1.61 are grams of water used

CaO (s) +H2O (l) --> Ca(OH)2 (s) 

# of moles of CaO and H2O 
n = given mass / molar mass. 
Ca = 40 
O = 16 
Mass CaO = 56 grams/mole 

n = 2.50 / 56
grams/mole = 0.0446 moles 

Molar mass = 2*H + O = 2 + 16 = 18 grams/ mol 

n = 2.41 / 18 = 0.134 (WATER) 

Now,
The number of moles of water left over. 

moles of water - moles of water used in the reaction = left over moles. 
0.134 - 0.0446 = 0.0893 

The grams of water left over. 

1 mole of water = 18 grams 
0.0893 moles of water = X

X = 18 x 0.0893 = 1.61 grams of water not used in the reaction.

Ireally don’t know this