How do you do limiting reactant, theoretical yield and precent yield problems?

Question

Like how do you do this question:

When 36.8 g of C6H6 react with 36.9 g of chlorine, the actual yeild of C6H5Cl is 38.8 g. What is the precent yield of chlorobenzene?

1. Determine limiting reactant

2. Use limiting reactant to calculate the theoretical yield

3. Find the percent yield.

Answer 1

The best way to do limiting reagent (reactant) problems is to do the stoich for both reactants, then analyze the results.  I split it into 3 sections.  2C6H6 + Cl2 --> 2C6H5Cl

I - do the stoich for both

36.8 g C6H6 will produce 48.4 grams of product

36.9 g Cl2 will produce  107 g of product

II the smaller amount is the expected yield or theoretical yield, so you get 48.4 g of product and C6H6 is the limiting reagent.

III I usually make them determine the excess reagent, so you would take (1-ratio of answers from part I) * original amount of excess reagent = leftover reagent

So for this problem (1-48.4/107) * 36.9 = 20.2 g excess

The percent yield is the actual/theoretical yield, so 38.8g/48.4 g = 80.2% which isn't great.  

Answer 2

Write the balanced equation-

C6H6 + Cl2 --> C6H5Cl + HCl

1 mol C6H6 reacts with 1 mole Cl2 to yield 1 mole of C6H5Cl

Calculate the moles of each reactant-

36.8 g C6H6 / 78.1 g/mol = 0.471 mol C6H6

36.9 g Cl2 / 70.9 g/mol = 0.520 mol Cl2

0.471 mole C6H6 requires 0.471 mole Cl2. You have excess Cl2 so C6H6 is limiting.

0.471 mole C6H6 will yiels (at 100%) 0.471 mole C6H5Cl

0.471 mol x 112.5 g/mol = 53.0 g C6H5Cl

38.8 g actual / 53.0 theoretical = 0.732 (73.2% yield)