The mass percentage of H 2 S in the sample has to be calculated.

The reactions are given below.

    Cd2++H2S→ CdSCdS+I2→ S+ CdI2I2+KI→KI32 S2O32−+I3−→I2+ S4O62−

Given that 7.65 mL (0.00765 L) of 0.0750 M Na2S2O3 is needed for the titration.

That is, number of moles of Na2S2O3 needed for titration is 5.7×10−4(0.00765 L×0.0750 M) mole.

Therefore according to the stoichiometry of the reaction number of moles of I3− reacted is 2.8×10−4(5.7×10−42) mole.

Number of moles of I2 will be equal to number of moles of I3−.

That is 2.8×10−4 mole of I2 was present in excess.

Given that 30 mL (0.030 L) of 0.0115 M I2 was added initially. That is  3.45×10−4(0.030 L× 0.0115 M)  moles of I2 was present initially.

Number of moles of I2 reacted will be 6