The amount of iodine obtained is 1.056×10^4 g
The balanced equation representing the formation of I2 from I03 and HSO 3is as follows: 2I03 (aq) + 5HSO3(aq) =;3HSO4 (aq) + 2SO4 (aq) + Hâ‚‚O(l) + I2 (s)
According to the equation 2 moles of IO3 produces 1 mole of I2. Molar mass of mass of iodine (i2) is 253.8 g/mol'
Relationships used are as follows: 2 mol IO3 = 1 mol Iâ‚‚1 mol I2= 253.8 g I2, Number of moles of NaIO, is 83.23 mol NaIO3.
Multiply the number of moles of NaIO3, with the conversion factors obtained from the mole ratio of the stoichiometric equation and the molar mass of I2 to calculate the amount of Iâ‚‚ produced.
(83.23 ×1×1×253.8)/1×2×1 = 1.056×10^4 g
Therefore, amount of iodine obtained is 1.056×10^4 g
Learn more about Iodine here:
https://brainly.com/question/16867213
#SPJ4
