In a sample consisting of 1.00 mol nabr and 0.300 mol ki, what is the mass percent of iodine? (a) 24.9% (b) 32.6% (c) 47.2% (d) 83.1

The quantities you have to know are the molar masses and the molecular stoichiometric ratios. The solution for this problem is as follows:

Mass Percent of Iodine = (Mass of Iodine/Mass of NaBr and KI)*100

Mass of NaBr = 1 mol NaBr * 102.89 g/mol = 102.89 g
Mass of KI = 0.3 mol KI * 166 g/mol = 49.8 g
Mass of KI = 0.3 mol KI * 1 mol I/1mol KI * 1 mol I₂/2 mol I * 253.81 g/mol = 38.07 g

Mass %I₂ = 38.07/(102.89+49.8) * 100 = 24.9%

Use a magnet to separate iron, evaporate water, i'm not sure how to separate the other two i'm sorry but good luck

The answers are 3.c kelvin4.b liquids only