The theoretical yield of I2 in the reaction would be 0.23 g
This refers to the stoichiometric yield of a reaction.
From the equation of the reaction:
Ca(IO3)2 + 10 KI + 12 HCl → 6 I2 + CaCl2 + 10 KCl + 6 H2O
The mole ratio of Ca(IO3)2 and I2 is 1: 6
Mole of 15.00 mL, 0.0100 M Ca(IO3)2 = 15/1000 x 0.0100
= 0.00015 mole
Equivalent mole of I2 = 0.00015 x 6
= 0.009 mole
mass of 0.0009 I2 = 0.0009 x 253.809
= 0.23 g
More on stoichiometric calculations can be found here: https://brainly.com/question/6907332
