A sample of hydrated cobalt (II) chloride has a mass of 5.22 g. After heating, it has a mass of 2.85 g. What is the percent by m
2 answers:
<u>Answer:</u> The percent by mass of water in the hydrate is 45.4 %
<u>Explanation:</u>
The chemical equation for the heating of hydrated cobalt (II) chloride follows:
We are given:
Mass of hydrated cobalt (II) chloride = 5.22 g
Mass of anhydrous cobalt (II) chloride = 2.85 g
Mass of water released = (5.22 - 2.85)g = 2.37 g
To calculate the mass percentage of water in the hydrate, we use the equation:
Mass of hydrated cobalt (II) chloride = 5.22 g
Mass of water = 2.37 g
Putting values in above equation, we get:
Hence, the percent by mass of water in the hydrate is 45.4 %
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Explanation:
Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.
This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.
- On addition of reactant at equilibrium shifts the equilibrium in forward direction.
- On addition of product at equilibrium shifts the equilibrium in backward direction.
- On removal of reactant at equilibrium shifts the equilibrium in backward direction.
- On removal of product at equilibrium shifts the equilibrium in forward direction.
Reactants = A , B
Product = C
1. Increase A
On increasing the amount of A at equilibrium will shift the equilibrium in forward or rightward direction.
2. Increase B
On increasing the amount of B at equilibrium will shift the equilibrium in forward or rightward direction.
3. Increase C
On increasing the amount of C at equilibrium will shift the equilibrium in backward or leftward direction.
4. Decease A
On decreasing the amount of A at equilibrium will shift the equilibrium in backward or leftward direction.
5. Decease B
On decreasing the amount of B at equilibrium will shift the equilibrium in backward or leftward direction.
6. Decease C
On decreasing the amount of C at equilibrium will shift the equilibrium in forward or rightward direction.
7. Double A and Halve B
Equilibrium constant of the reaction = K
On doubling A and halving B, equilibrium constant of the reaction = K'
The value of equilibrium constant K' is equal to K, which means that equilibrium will not shift in any direction.
8. Double both B and C
Equilibrium constant of the reaction = K
On doubling B and C, equilibrium constant of the reaction = K'
K' = 2 K
The value of equilibrium constant K' is double the K, which means that product is increasing which means that equilibrium will shift in backward or leftward direction.