See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

           2 2 H g I g 2HI g Initial mol 1 1 0 mol at eqbm 1 x 1 x 2x 1 x 1 x 2x eqbm conc V V V        Now, equivalents of hypo used = equivalents of I2 present at equilibrium    