The iodide in a sample that also contained chloride was converted to iodate by treatment with an excess of bromine: The unused bromine was removed by boiling; an excess of barium ion was then added to precipitate the iodate: In the analysis of a 1.54-g sample, 0.0596 g of barium iodate was recovered. Express the results of this analysis as percent potassium iodide.

The percentage of potassium iodide in the sample is 2.63 %.

Explanation:

The chemical equation for the reaction of iodide ions with bromine gas follows:

               (i)

The chemical equation for the reaction of iodate ions with barium ions follows:

                ......(ii)

To calculate the number of moles, we use the equation:

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

By Stoichiometry of the reaction (ii):

1 mole of barium iodate is produced by 2 moles of iodate ions

So, of barium iodate will be produced by of iodate ions

By the stoichiometry of the reaction (i):

1 mole of iodate ions are produced by 1 moles of iodine ions

So, of iodate ions will be produced by of iodine ions

Moles of potassium iodide = Moles of iodide ions =

Since, the molar mass of potassium iodide = 166 g/mol

Using equation 1:

To calculate the percentage by mass of a substance, the equation used is:

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

Hence, the percentage of potassium iodide in the sample is 2.63 %.

3

explanation:

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