When fec13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4fecl3(sj 30lgj ~ 2f~0 (sj 6cl2(gj if 3.00 mol of fec13 are ignited in the presence of 2.00 mol of 0 2 gas, how much of which reagent is present in excess and therefore remains unreacted

Answer : The reagent present in excess and remains unreacted is,

Solution : Given,

Moles of = 3.00 mole

Moles of = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

From the balanced reaction we conclude that

As, 2 moles of react with 1 mole of

So, 3.00 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is,

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%

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explanation: