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.

I’d have to say no because without oxygen in the water the fish would all be dead. so if you want to fish for dead fish, pick the ponds with the lowest oxygen concentrations.

Variables,control,conclusion and finally scientific theory