in an experiment 5.00ml of 0.1440 mol L^1- aqueos KIO3 was reacted with an excess of aqueous iodide and the molecular iodine I2, Formed was titrated against 0.1000mol L^1- aqueous S2O3^2- the equation for the titration reaction is:

I2(aq)+ 2 S2O3^2-(aq)---->S4O6^2-(aq)+ 2I^-1(aq)

calculate the number of moles of potassium iodate (KIO) that were present in the 5.00mL aliquot of standar potassium iodate solution

can some1 please help me?

I've answered this for you earlier tonight. moles KIO3 = M x L = 0.144M x 0.005 mL = ?

To determine the number of moles of potassium iodate (KIO3) present in the 5.00 mL aliquot of standard potassium iodate solution, you need to use stoichiometry and the balanced equation for the titration reaction.

Given:
Volume of KIO3 solution (V1) = 5.00 mL = 0.00500 L
Molarity of KIO3 solution (M1) = 0.1440 mol L^-1

From the balanced equation:
1 mole of KIO3 reacts with 1 mole of I2

First, let's calculate the number of moles of I2 that reacted in the titration using the molarity and volume of the thiosulfate solution:

Molarity of thiosulfate solution (M2) = 0.1000 mol L^-1

Using the equation M1V1 = M2V2:
(0.1440 mol L^-1)(0.00500 L) = (0.1000 mol L^-1)(V2)
V2 = (0.1440 mol L^-1)(0.00500 L) / (0.1000 mol L^-1)
V2 = 0.00720 L

This means that you used 0.00720 L of the thiosulfate solution in the titration.

Using the balanced equation, we know that 1 mole of I2 reacts with 2 moles of thiosulfate ions (S2O3^2-).

Therefore, the number of moles of I2 would be half the number of moles of thiosulfate ions used in the titration:

Number of moles of I2 = (0.1000 mol L^-1)(0.00720 L) × 1/2
Number of moles of I2 = 0.000360 mol

Since 1 mole of KIO3 reacts with 1 mole of I2, the number of moles of KIO3 in the original aliquot of KIO3 solution is equal to the number of moles of I2:

Number of moles of KIO3 = 0.000360 mol

Therefore, there are 0.000360 moles of KIO3 present in the 5.00 mL aliquot of standard potassium iodate solution.