# chemistry

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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)

determine the molar ratio of iodine molecules by dividing each mole quantity by the lowest mole quantity. express this as a ratio of integers

IO3- : I2

using this ratio balance the equation please

__IO3^-(aq) + ____ I^-(aq)+____H^+(aq)------>_____ I2(aq)+_____H2O(l)

• chemistry -

I would do it this way.
moles KIO3 = M x L = 0.144 x 0.005 = 0.00072 moles KIO3.
moles S2O3^2- = M x L = 0.1000 x 0.043 = 0.0043 moles S2O3^2-. Convert that to moles I2.
moles I2 = 0.0043 moles S2O3^2- x (1 mole I2/2 moles S2O3^-) = 0.0043 x (1/2) - 0.00215 moles.
So the ratio IO3^- to I2 = 0.00072:0.00215.
To find the ratio in small whole numbers, the easy way to do it is to divide the smaller number by itself, then divide the other number by the same value.
IO3 = 0.00072/0.00072 = 1.000
I2 = 0.00215/0.00072 = 2.986 which rounds in whole numbers to 3.000. (The difference is in your titration. That may have been slightly different than exactly 43.0 mL).
Knowing 1 IO3 = 3 I2, put 1 for the coefficient for IO3, place 3 for I2, then balance the remainder of the equation by inspection. You should get
IO3^- + 5I^- + 6H^+ ==> 3I2 + 3H2O