Asked by Beth
An anyhdrous metal chloride, X, contains 37.4% Cl. X is a reducing agent, 0.895 g requiring 23.60 cm^3 of 0.100M KIO3 for complete oxidation in 3- 9M HCl. X reacts with potassium chloride to form an adduct Y, containing 14.8% K. Identify X and Y, and comment briefly on their stereochemistry. Write equations for all reactions and state how you would determine the endpoint of the iodate titration.
I worked out X to be SnCl2 and found the reaction ratio with KIO3 to be 2:1, so the equation is 2snCl2 + KIO3 + 6H+ + 6Cl- --> KICl2 + 2SnCl4 + 3H2O
I then found Y to be KSnCl3
I don't know what stereochemistry means or what to write about X and Y, and I don't know about the endpoint. Is what I have done so far correct?
I worked out X to be SnCl2 and found the reaction ratio with KIO3 to be 2:1, so the equation is 2snCl2 + KIO3 + 6H+ + 6Cl- --> KICl2 + 2SnCl4 + 3H2O
I then found Y to be KSnCl3
I don't know what stereochemistry means or what to write about X and Y, and I don't know about the endpoint. Is what I have done so far correct?
Answers
Answered by
DrBob222
Yes, I think you are ok. I'm almost positive that compound X is SnCl2 and compound Y is KSnCl3. The percentages work out ok for those two compounds, too. Stereochemistry means the spatial arrangement. The best evidence I can find points to a planar arrangement for SnCl2 and a pyramidal arrangement for the SnCl3^- ion (the KSnCl3 compound probably is ionic). As for the titration, I would think along the lines of potentiometric or polarographic. I am dubious of the equation you have written; what evidence do you have that the IO3^- is reduced to the IO^- ion and not to I2. I would have expected IO3^- to be reduced to I2 and Sn+2 to be oxidized to Sn^+4; however, that doesn't give the correct ratio for the moles IO3^- to mols Sn^+2 to be
SnCl2 for compound X.
SnCl2 for compound X.
Answered by
Beth
I am not sure about the equation either. Obviously if I2 was being produced in the titration then the end point could be tested with starch, but the compounds I have found seem to be right according to the percentages, so I'm not sure what else the equation could be.
Answered by
Marylee
...hmmmm are you a Durham Chemistry Student?
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