Cao + H2O ---->Ca(OH)2
Calculate the number of each reactant and the moles of the product
that is
moles = mass/molar mass
The moles of CaO= 56.08g/ 56.08g/mol(molar mass of Cao)= 1mole
the moles of water= 36.04 g/18 g/mol= 2.002moles
The moles of Ca (OH)2=74.10g/74.093g/mol= 1mole
The mass of differences of reactant and product can be therefore
explained as
1 mole of Cao reacted completely with 1 mole H2O to produce 1 mole of Ca(OH)2. The mass of water was in excess while that of CaO was limited
C. The reaction goes from one of less moles to more causing it to be more disordered and therefore have a positive change in s
Answer: The concentrations of
at equilibrium is 0.023 M
Explanation:
Moles of
= 
Volume of solution = 1 L
Initial concentration of
= 
The given balanced equilibrium reaction is,

Initial conc. 0.14 M 0 M 0M
At eqm. conc. (0.14-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO%5D%5Ctimes%20%5BCl_2%5D%7D%7B%5BCOCl_2%5D%7D)
Now put all the given values in this expression, we get :

By solving the term 'x', we get :
x = 0.023 M
Thus, the concentrations of
at equilibrium is 0.023 M
Answer:
D. 5 moles
Explanation:
C3H8 + 5O2 → 4H2O + 3CO2
5 mol 3 mol
So, to make 3 mol CO2 5 mol O2 are needed.