Given:

Concentration of Na2S2O3 is 0.1979 M

Concentration of KIO3 is 0.1201 M

Volume of 0.5 M of KI is 0.01 L

Volume of 1.0 M HCL is 0.01 L

Volume of KIO3 = 0.005 L

Average Volume of Na2S2O3 is 0.0184 L

(2.1) IO3 (aq) + I (aq) + H (aq) ----> I2(aq) + H2O(l)

(2.2) mole of I total = mole of IO3- +mole of I-

(2.3) I2(aq) + 2S2O3 (aq) ----> S4O6 (aq) + 2I (aq)

1. Calculate the number of moles of Na2S2O3 in the average volume of Na2S2O3. Using equation (2.3) calculate the number of moles of molecular iodine. This is the same as the number of moles of molecular iodine produced by reaction (2.1).

2. All of the iodine produced by reaction (2.1) is in the form of I2 (molecular iodine). Calculate the number of moles of atomic iodine (I) produced by reaction (2.1).

3. Calculate the number of moles of KIO3 consumed by reaction (2.1). Remember, it is the limiting reagent.

4. Calculate the number of moles of I- consumed by reaction (2.1).

5. Determine the ratio for IO3 : I : I2 for reaction (2.1). Express this as a ratio of integers.

6. Using the results for question 5 complete the balancing of: ____IO3- (aq) + ____I- (aq) + ____H+ (aq) ---> ____I2(aq) + ____H2O(l)

Answer rating: 100% (QA)

1 moles M x V moles of Na 2 S 2 O 3 01979 M x 00184 L 00036 moles of Iodine 1 mole of I 2 2 mole of ... View the full answer