Question

1. What is the theoretical
 yield of C6H5Br in this reaction when 30.0 g of C6H6
reacts with excess Br2? If the 
actual yield of C6H5Br was 56.7 
g, 
what 
is
 the 
percent 
yield?

Answer 1

Answer:

The theoretical yield in grams is 60.292 grams, and in moles the theoretical yield is 0.384. The percent yield is 94%.

Explanation:

Answer 2

The theoretical yield of bromobenzene is 60.3 grams and the percent yield of the reaction is 94.0%.

Explanation:

• The percentage yield of the reaction is given by :

$Yield(\%)=\frac{\text{Actual or experimental yield}}{\text{Theoretcial yield}}\times 100$

• Experimental yield is the amount of product produced after the carrying out the reaction practically.

• Theoretical yield is the amount of product calculated mathematically by using the amount of reactant.

Given:

Mass of benzene = 30.0 g

The actual yield of bromobenzene = 56.7 g

To find:

Theoretical yield and percent yield of the reaction.

Solution:

Mass of benzene = 30.0 g

Moles of benzene =$\frac{30.0 g}{78.11 g/mol}=0.384 mol$

$C_6H_6+Br_2\rightarrow C_6H_5Br+HBr$

According to reaction, 1 mole of benzene gives 1 mole of bromobenzene, then 0.384 moles of benzene will give:

$=\frac{1}{1}\times 0.384 mol=0.384 \text{mol of bromobenzene}$

Mass of 0.384 moles bromobenzene:

$=0.384 mol\times 157.02g/mol=60.3 g$

The theoretical yield of bromobenzene = 60.3 g

The actual yield of bromobenzene = 56.7 g

The percent yield of the reaction:

$Yield(\%)=\frac{56.7 g}{60.3 g}\times 100=94.0\%$

The theoretical yield of bromobenzene is 60.3 grams and the percent yield of the reaction is 94.0%.

Learn more about the percent yield of reaction here:

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