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Preparation and Composition of Tin (IV) Iodide

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Introduction

Preparation and Composition of Tin (IV) Iodide Procedure:- > 3.0020g of Tin (II) Chloride (SnCl2.2H2O) was mixed with 25 cm3 of an equivolume mixture of glacial acetic acid and acetic anhydride in a 100 ml "Quickfit" flask. > In order to dehydrate the Tin (II) Chloride, the mixture was boiled under reflux for 15 minutes. > The solution and the precipitated Tin (II) Chloride were allowed to cool a little. > And then, while shaking the flask, 3.2950g of iodine was added in portions down the condenser. > A cream-coloured precipitate was formed which was dichloro-di iodo tin (SnCl2I2). This redissolved as the last portions of iodine were added. > Then the solution was refluxed for a few minutes to complete the reaction and was allowed to cool. > Tin (IV) iodide came down as small orange crystals while the chloride remained in the solution. > The solid was filtered off rapidly by suction. ...read more.

Middle

x 10 -2 moles of SnCl2I2 Equation 2: 2SnCl2I2 -----------------> SnCl4 + SnI4 From equation (2); 2 moles SnCl2I2 produces 1 mole of SnI4 Therefore, 1.306 x 10 -2 moles of SnCl2I2 should produce x 1.306 x 10 -2 moles of SnI4 = 6.53 x 10 -3 moles of SnI4 Mass of 6.53 x 10 -3 moles of SnI4 ( Theoretical yield) Mr of SnI4 = 626.3286 Mass = (6.53 x 10 -3) X 626.3286 = 4.0899g Actual yield = 2.0850g % yield = X 100 = 51 % Percentage of Iodine in Tin (IV) Iodide Procedure:- > Approximately 0.25g of SnI4 was weighed in a glass stoppered flask, 30cm3 of concentrated HCl and a few cm3 of chloroform were added to it. > The solution was then titrated with a standard solution of potassium iodate ( containing about 1.25g in 250 cm3) until on vigorously shaking of the flask, the violet coloration in the organic layer just disappeared ( i.e the aqueous part becomes pale yellow). ...read more.

Conclusion

Calculating % of iodine in the sample Average mass of sample used = = 0.25185g Average volume of potassium iodate used = = 34.95 cm3 250cm3 of KIO3 solution contains 1.25g of potassium iodate. Therefore, 34.95cm3 of KIO3 solution contains X 34.95 g of potassium iodate = 0.17475g of potassium iodate Number of moles of KIO3 in 0.17475g of KIO3 Mr of KIO3 = 214.0011 Number of moles = = 8.166 x 10 -4 moles Equation: IO3- + 2I- + 6H+ + 5Cl- ----------> 5ICl + 3H2O From the equation, 1 mole of IO3- reacts with 2 moles of iodide Therefore, 8.166 x 10 -4 moles of IO3- reacted with 2 x (8.166 x 10 -4 moles) of iodide. =1.6332 x 10-3 moles of iodide. Mass of iodide present in the sample Ar of iodine = 126.9045 Mass of iodide= Ar x number of moles = 126.9045 x 1.6332 x 10-3 = 0.20726g Percentage of iodine in the sample Mass of sample= 0.25185g Mass of iodide= 0.20726g % of iodine in sample = x 100 =82.30% ?? ?? ?? ?? Page 1 ...read more.

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