Aqueous potassium iodate and potassium iodide react in the presence of dilute hydrochloric acid, as shown below. kio3(aq) + 5ki(aq) + 6hcl(aq) ⟶ 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.

0.0797 g Looking at the formula, 1 mole of KIO3 and 5 moles of KI will react and produce moles of iodine molecules or 6 moles of iodine atoms. So first, determine the number of moles of KIO3 and KI provided moles KIO3 = 0.0121 * 0.097 = 0.0011737 mol moles KI = 0.0308 * 0.017 = 0.0005236 mol The limiting reactant is KI at 0.0005236 mol so divide by 5 and multiply by 6 to get the number of moles of iodine atoms. 0.0005236 / 5 * 6 = 0.00062832 mol Lookup the atomic weight of iodine which is 126.90447 And multiply that by the number of moles of iodine produced 126.90447 g/mol * 0.00062832 mol = 0.079736617 g Rounding to 4 decimal places gives 0.0797 g

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Ido not understand.