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Question

Iodine in a sample containing iodine and chlorine ion is converted to the iodate ion adding excess amount of bromine.

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3H2O +3Br2 + I - <-----> 6Br‾ + IO3‾ +6H+

Then remaining bromine is removed by boiling and iodate ions are precepitated adding excess barium ion.

Ba2+ + 2IO3‾ <------> Ba(IO3)²

In analysis of 2.72 g of sample, 0.0720 g of barium iodate (487.13 g/mol) is obtained. Express the result of analysis in % of KI (166.0 g/mol).

PLEASE WRITE THE SOLUTION WITH STEPS

Answer 1

You are asked for the % of KI, which means we need to find the mass of KI present in the original sample. Once we have that, we simply divide it by the mass of the entire sample (2.72 g) to find % KI.

How can we find the mass of KI? We use the stoichiometry of the balanced equations along with dimensional analysis:

0.0720 g Ba(IO₃)₂ x 1 mol Ba(IO₃)₂ / 487.13 g x 2 mol IO₃^- / mol Ba(IO₃)₂ = 0.0002956 moles of IO₃^-

Looking at the first reaction, we see that 1 mol IO₃^- comes from 1 mole I^- (or 1 mol KI to be exact). So...

moles of I^- = moles KI = 0.0002956 mol IO₃^- x 1 mol I^- / mol IO₃^- = 0.0002956 moles I^- = 0.0002956 moles KI

mass KI = 0.0002956 moles KI x 166.0 g/mol = 0.0491 g KI in original sample

%(m/m) KI = 0.0491 g / 2.72 g (x100) = 1.80% KI