Question

Iodine is made by the following reaction 2 NaIO3(aq) + 5 NaHSO3(aq) → 3 NaHSO4(aq) + + 2 Na2SO4(aq) + H2O(ℓ) + I2(aq) If you wish to prepare 2.65 kg of I2, what masses of NaIO3 and NaHSO3 are required?

____g NaIO3

____g NaHSO3

Iodine is made by the following reaction 2 NaIO3(aq) + 5 NaHSO3(aq) → 3 NaHSO4(aq) + + 2 Na2SO4(aq) + H2O(ℓ) + I2(aq) What is the theoretical yield of I2 if you mixed 12.0 g of NaIO3 with 195 mL of 0.853 M NaHSO3?

_____g I2

Answer 1

2NaIO3 + 5NaHSO3 --------------> 3NaHSO4 + 2N2SO4 + H2O + I2

according to balanced reaction

253.809 g I2 require 2 x 197.892 g NaIO3

2650 g (2.65 kg) I2 require 2650 x 2 x 197.892 / 253.809 =4132.35 g

mass of NaIO3 required = 4132.35 g

now

253.809 g I2 require 5 x 104.061 g NaHSO3

2650 g I2 require 2650 x 5 x 104.061 / 253.809 = 5432.46 g

mass of NaHSO3 require 5432.46 g

part 2)

according to balanced reaction

2 x 197.892 g NaIO3 reacts with5 moles NaHSO3

12.0 g NaIO3 reacts with 12.0 x 5 / 2 x 197.892 = 60/ 395.784 = 0.15 moles

now calculate the moles of NaHSO3 given

moles of NaHSO3 = 0.853 x 195/1000 = 0.166 moles

so NaHSO3 is in exess

NaIO3 is limiting reagent.

now

2 x 197.892 g NaIO3 gives 253.809 g I2

12.0 g NaIO3 gives 12.0 x 253.809 / 2 x 197.892 = 3045.708 / 395.784 = 7.70 g

therotical yield of I2 = 7.70 g