Answer:
1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2
Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.
Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.
The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.
Explanation:
the total volume of the mixture is equal to:
Vmix = 75 + 30 + 25 + 5 + 5 + 10 = 150 mL
the moles of each species in the mix equals:
Na2S2O3 = 0.05 * 0.005 = 0.00025 moles
KI = 0.05 * 0.025 = 0.00125 moles
H2O2 = 1.02 * 0.01 = 0.0102 moles
the following equation shows the reaction between I2 and S2O32:
I2 + S2O32 = 2I- + S4O62-
The same way:
2I- + 2H+ + H2O2 = I2 + 2H2O
1 mol of H2O2 consumes 1 mol of S2O32, then 0.0005 mol of I- reacts with 0.00025 moles of H2O2
Thus, initial moles of H2O2 present in the mix is equal to 0.0102 moles.
Amount of moles of H2O2 that reacts with I- is equal to 0.00025 moles.
The amount of I ions that is oxidized by H2O2 and the iodine, I2 that is produced, is consumed by the S2O32- ions and are transformed into I ions. The amount of I ions that are consumed by H2O2 is equal to the amount of ion that is produced by S2O32-.
