Calcium reacts with oxygen to make calcium oxide, CaO. What is the theoretical yield (g) of CaO(s) that could be prepared from 7.75 g of Ca(s) and 2.44 g of O2(g)

Explanation:

2Ca    +    O₂   =    2CaO

80 g     32 g         112 g

given amount of Ca = 7.75 g

it will require oxygen =  32 x 7.75 / 80 = 3.1 g

but available oxygen = 2.44 so oxygen is the limiting reagent .

32g oxygen reacts with calcium to give 112 g of calcium oxide

2.44 gram of oxygen reacts with calcium to give 112 x 2.44 / 32 of CaO .

= 8.54 gram .

Theoretical yield of CaO = 8.54 gram.

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%