Table I. Composition of Reaction Mixtures Solution A Solution B Buffer: Mixture

Question

Table I. Composition of Reaction Mixtures Solution A Solution B Buffer: Mixture of 0.020 MStarch Solution Na2S203 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL Trial 0.5MCHCo,H & DI Water 0.30 M KI 0.10 MH202 0.5 M NaCH CO 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 1.0 mL 5 dro 5 dro 5 dro 5 dro 5 dro 5 drops 5 dro 5 dro 5 dro 5 dro 5 dro 5 drops 5 dro 6.0 mL 5.0 mL 4.0 mL 3.0 mL 2.0 mL 1.0 mL 7.0 mL 5.5 mL 4.0 mL 3.0 mL 1.5 mL none 7.0 mL 1.0 mL 2.0 mL 3.0 mL 4.0 mL 5.0 mL 6.0 mL 2.0 mL 2.0 mL 2.0 mL 2.0 mL 2.0 mL 2.0 mL 1.0 mL 3.0 mL 3.0 mL 3.0 mL 3.0 mL 3.0 mL 3.0 mL 1.0 mL 2.5 mL 4.0 mL 5.0 mL 6.5 mL 8.0 mL 3.0 mL 2 3 4 10 12 none

Explanation / Answer

The given reaction is Equation: 2I-+H2O2 + 2 H3O+ ----> I2 +4H2O     equation 1
CLOCK reaction: 2S2O32- +I2 ----> 2I- + S4O62-      equation 2

we calculate the number of moles contains by solution B .

solution B have 3 ml of 0.10 H2O2

molarity of H2O2 =    3 ml* (1lit   /1000 ml) * 0.10 mole / lit = 0.0003 mole

from the reaction we can say one mole of H2O2 react with 2 mole of I-

giving 1mole I2 (as seen in equation 1)

moles of I- =  1ml * (1 lit/ 1000 ml) 0.30 mol /lit = 0.0003 mole

so from the reaction we can say one mole of H2O2 reacts with 2 mole of I- to give 1 mole of I2

here I- is our limiting reagent .

in the next reaction 1 mole of I2 react with 2 moles of S2O32- .

so 0.0001 moles of I- react only with 0.0002 moles of S2O32- .

So the moles of S2O32-consumed in trial 1 = 0.0002 moles .

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