Can somebody help me with this percentage yield problem?

Question

For the reaction represented by the equation Cl2 + 2KBr => 2KCl + Br2, calculate the percentage yield if 200. g of chlorine react with excess potassium bromide to produce 410. g of bromine. (Cl = 35.5 amu; Br = 79.9 amu).

 

It would help if you explained the steps.

Thanks!

 

Answer 1

Hi Sarah,

 

A percent yield problem requires you to take the actual yield of product and divide it by the theoretical yield, which is based on the limiting reactant. In this problem, you are told that KBr is in excess, so you know Cl₂ is the limiting reactant. You can proceed from there.

 

Since you are given the amount of one substance (Cl₂) and are required to calculate the amount of a different substance (Br₂), the first part of this problem is a stoichiometry problem, and all stoichiometry problems need to go through moles. I always tell my students (and sometimes they even listen) to let the units be your guide when setting up one of these calculations, so (units only, no numbers yet):

 

(__g Cl2) × (1 mol/__g Cl₂) × (__mol Br₂/__mol Cl₂) × (__g Br2/1 mol Br2) = ___ g Br2 (theoretical yield)

 

The values are taken from (in order) (1) given mass of Cl₂ (2) molar mass of Cl₂ (3) mole to mole ration from the balanced equation (4) molar mass of Br₂. Plugging in the appropriate numbers and punching it into your calculator will give you the theoretical yield of Br₂.

 

The percent yield is simply (actual yield of Br₂)/(theoretical yield of Br₂) × 100.

 

Jim Scripko