Preparation and Composition of Tin (IV) Iodide

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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. (Tin (IV) iodide is hydrolysed by the moisture in the air)
  • The product was weighed in a well fitting stopper and the results tabulated.

Table of results

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Calculation of the % yield

Mr of SnCl2.2H2O = 229.9350

Mr of SnCl2 = 189.616

Mass of SnCl2.2H2O weighed = 3.0020g

Mass of SnCl2 in 3.0020g of SnCl2.2H2O =  x 3.0020g

                                             = 2.4756g of SnCl2

Therefore, this mass (2.4756g) of SnCl2 reacted with the 3.2950g of iodine to give SnCl2I2.

                               Equation 1:             SnCl2   +  I2  ------------>  SnCl2I2

Number of moles of SnCl2 =

                                = 1.306 x 10 -2 moles ...

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