Chemistry: 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.

Igonna be honest i dunno that one srry...

The iodide in a sample that also contained chloride

natasniebow

06.05.2021

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:

$I^-+3Br_2+3H_2O\rightarrow 6Br^-+IO_3^-+6H^+$ (i)

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

$Ba^{2+}+2IO_3^-\rightarrow Ba(IO_3)_2$ ......(ii)

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of barium iodate = 0.0596 g

Molar mass of barium iodate = 487.13 g/mol

Using equation 1:

$\text{Moles of barium iodate}=\frac{0.0596 g}{487.13 g/mol}\\\\\text{Moles of barium iodate}=1.22\times 10^{-4} moles$

By Stoichiometry of the reaction (ii):

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

So, $1.22\times 10^{-4} moles$ of barium iodate will be produced by $\frac{2}{1}\times 1.22\times 10^{-4} =2.44\times 10^{-4}moles$ of iodate ions

By the stoichiometry of the reaction (i):

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

So, $2.44\times 10^{-4} moles$ of iodate ions will be produced by $\frac{1}{1}\times 2.44\times 10^{-4} =2.44\times 10^{-4}moles$ of iodine ions

Moles of potassium iodide = Moles of iodide ions = $2.44\times 10^{-4}$

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

Using equation 1:

$\text{Mass of potassium iodide}=2.44\times 10^{-4}mol\times 166 g/mol\\\\\text{Mass of potassium iodide}=0.0405 g$

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

$\text{Percent by mass}=\frac{\text{Mass of a substance}}{\text{Mass of solution}}\times 100$

Mass of a solution = 1.54 g

Mass of potassium iodide = 0.0405 g

Using above equation:

$\text{Percent potassium iodide}=\frac{0.0405 g}{1.54g}\times 100\\\\\text{Percent potassium iodide}=2.63\%$

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

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