Question

Using the equation c6h6 + cl2 = c6h5cl + hcl what is the theoretical yield of c6h5cl if 91.2 of c6h6

Answer 1

The theoretical yield of a reaction is the amount of a product produced if all of the reactant is consumed in the reaction or the conversion is 100%. We determine this value by using the initial amount of the reactant and the relation from the chemical reaction. We do as follows:

91.2 g C6H6 (1 mol C6H6 / 78.12 g C6H6) (1 mol C6H5Cl / 1 mol C6H6) ( 112.56 g / 1 mol ) = 131.41 g C6H5Cl

Answer 2

Answer:

Theoretical yield of $C_{6} H_{5} Cl$ is 53 g and percentage yield is 73.2% after chlorine react in order to produce chlorobenzene.

Concept:

The amount of a product that results from the total conversion of the limiting reactant in a chemical process is known as the theoretical yield. It is not the same as the quantity you would actually receive from a reaction in the lab because it represents the amount of product that would occur from a flawless (theoretical) chemical reaction. Common ways to express theoretical yield are in terms of grams or moles.

$Percentage yield = \frac{Actual yield}{Theoritical yield} *100$

Given:

Asprin is made using the chemical chlorobenzene, $C_{6} H_{5} Cl$. Benzene and chlorine need to react in order to produce chlorobenzene.

$C_{6} H_{6} + Cl_{2} \rightarrow C_{6} H_{6}Cl + HCl$

Find:

Theoretical yield of $C_{6} H_{5} Cl$.

Solution:

It is necessary to know the mole ratio of $C_{6} H_{6}$ to $C_{6} H_{5} Cl$ as well as their molecular masses.

$\frac{C_{6} H_{6} }{1} * \frac{1 mol C_{6} H_{6} }{78.12g C_{6} H_{6} } * \frac{1 mol C_{6} H_{5}Cl }{1 mol C_{6} H_{6} } * \frac{112.56g C_{6} H_{5}Cl }{1 mol C_{6} H_{5}Cl } =$ 53 g of $C_{6} H_{5} Cl$

So, theoretical yield = 53 g of $C_{6} H_{5} Cl$

Percentage yield = $\frac{38.8}{53} * 100 =$ 73.2%

So,  theoretical yield of $C_{6} H_{5} Cl$ is 53 g and percentage yield is 73.2%.

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