Question

What is my theoretical yield (in moles) of Potassium Bromide (KBr) if I start with 40 grams of Iron (II) Bromide [FeBr2]? moles KBr

Answer 1

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