In following reaction 3A(g)+B(g)⇌2X(g)+Y(g) if initial concentration of B is double that of A  and at equilibrium concentrations of A & X are equal the percentage B of reacted to attain equilibrium is

[B]initial =2[A]initid                                            3A   +   B   ⇌   2X+Y Initial concentration              c           2c          0      0                                         -3cα    -cα  +  2cα +cα Final concentration           c-3cα  2c-cα  2cα  cα ∵[A]equilibrium =[X]equilibrium  c-3cα=2cα5cα=cα=0.2 % of B reacted  =0.2c2c×100 =10.00