Question

Aqueous potassium iodate and potassium iodide react in the presence of dilute hydrochloric acid, as shown below. KIO3(aq) + 5KI(aq) + 6HCl(aq) \longrightarrow⟶ 3I2(aq) + 6KCl(aq) + 3H2O(l) What mass (in g) of iodine is formed when 12.1 mL of 0.097 M KIO3 solution reacts with 30.8 mL of 0.017 M KI solution in the presence of excess HCl? Enter to 4 decimal places.

Answer 1

molarity of KIO3 = number of moles of KIO3 / volume of solution in L

0.097 =  number of moles of KIO3 / 0.0121 L

number of moles of KIO3 = 0.00117 mole

molarity of KI = number of moles of KI / volume of solution in L

0.017 =  number of moles of KIO3 / 0.0308 L

number of moles of KI = 0.000524 mole

from the balanced equation we can say that

1 mole of KIO3 requires 5 mole of KI so

0.00117 mole of KIO3 willr require 0.00585 mole of KI

but we have 0.000524 mole of KI so KI is limiting reactant

from the balanced equation we can say that

5 mole of KI produces 3 mole of I2 so

0.000524 mole of KI will produce 0.000314 mole of I2

1 mole of I2 = 253.809 g

0.000314 mole of I2 = 0.0797g

Therefore , the mass of I2 produced will be 0.07974 g