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

Question

Armani Tyler

Chemistry

9 March, 03:33

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

+1

Answers (1)

Know the Answer?

Dania Bender · 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