Question

For the reaction A+3B=2C+D initial mole of A is twice as that of B .if at equilibrium moles of B &C are equal then %age of B reacted ?

Answer 1

For given reaction

                                    A       +         3B        ⇆       2C      +    D

Initial moles                2x                  x                      0             0

At equilibrium.         (2x-2y)           (x-y)                   2y            y

According to  question,

Moles of B and C are equal at equilibrium

Therefore,

x-y=2y

3y=x

Moles of B reacted at equilibrium = (x-y) = 3y-y = 2y

if x =1 then y = 0.3333

Thus,

Moles of B = 2×0.3333 = 0.6666

% of B reacted = 0.6666×100 = 66.66  

Answer 2

Answer:

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Explanation: