MY NOTES ASK YOUR TEACHER Consider a different titrant for this exercise. Suppose Ca(OH)2 were used as the titrant, instead of NaOH. This will make the titrant twice as concentrated in hydroxide ion. The analyte will still be HC2H3O2. (a) What is the stoichiometry of HC2H3O2 to Ca(OH)2

The stoichiometry of the reaction is 1:2.

Stoichiometry shows mass - mole relationship. Mass - mole relationships are very important in chemical arithmetic. All calculations in stoichiometry are based on the balanced chemical reaction equation.

The reaction equation for the reaction between Ca(OH)2 and HC2H3O2 is shown as follows; Ca(OH)2(aq) + 2HC2H3O2(aq) > Ca(C2H3O2)2(aq) + 2H2O(l). The stoichiometry of the reaction is 1:2.

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The stoichiometry of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2

To obtain the stoichiometry of HC₂H₃O₂ to Ca(OH)₂, we shall write the equation for the reaction. This is given below:

The above equation can be balanced as follow:

There are 2 atoms of C on the left side and 4 atoms on the right side. It can be balanced by writing 2 before HC₂H₃O₂ as shown below:

2HC₂H₃O₂ + Ca(OH)₂ —> Ca(C₂H₃O₂)₂ + H₂O

There are a total of 10 atoms of H on the left side and a total of 8 atoms on the right side. It can be balanced by writing 2 before H₂O as shown below:

2HC₂H₃O₂ + Ca(OH)₂ —> Ca(C₂H₃O₂)₂ + 2H₂O

Now, the equation is balanced.

From the balanced equation above, the mole ratio of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2

Therefore, the stoichiometry of HC₂H₃O₂ to Ca(OH)₂ is 1 : 2

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here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%