Considering the volumes and the concentrations used in mixture 1, what percentage of the moles of H2O2 present have been consumed during the timed reaction? Calculate the number of moles of H2O2 present initially and determine how many have reacted with iodide when the thiosulfate runs out. How does the I- concentration change during this same time?

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-.