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

Question

goldfiish [28.3K]

4 years ago

13

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

KBr

Chemistry

1 answer:

Verizon [17] · Answer 1 · 2021-11-17T08:06:58+03:00

Verizon [17]4 years ago

4 0

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