I start with 67.8 g of NaBr and excess chlorine gas. What is my theoretical yield of NaCl? b. I end up with a percent yield of 82.30%. What was the actual yield of NaCl?

Question

NaBr+Cl2=Br2+NaCl

partial answer: 3X.X g NaCl and 3X.X g NaCl

Answer 1

Answer 2

First, we need a balanced reaction.

2NaBr + Cl₂ ---> 2NaCl

The molar mass of NaBr = 22.98 + 79.90 = 102.88 g/mol, and for NaCl it's 22.98 + 35.45 = 58.43 g/mol.

67.8 g NaBr / 102.88 g/mol = 0.659 mol NaBr

Since the coefficients of both compounds each = 2, the theoretical yield of NaCl = 0.659 X 58.43 =

38.5 g NaCl.

With a percent yield of 82.30%, the actual yield was 38.5 g X 0.8230 = 31.7 g NaCl, rounded to 3 significant figures as is appropriate and suggested.