The reaction will be: FeBr2 + K --> KBr + Fe Balancing gives: FeBr2 + 2K --> 2KBr + Fe The molar mass of FeBr2 is 55.85 + 2*79.9 = 215.65 g/mol. We divide 40 g / 215.65 g/mol = 0.185 mol FeBr2 Based on stoichiometry: (0.185 mol FeBr2)(2 mol KBr/1 mol FeBr2) = 0.370 mol KBr
