Question

Consider the reaction of C6H6 + Br2 C6H5Br + HBr a. What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?

Answer 1

Answer:

this is a correct answer

Answer 2

The theoretical yield of C6H5Br if 42.1 g of C6H6 reacts with 73.0 g of Br2 is 72.2 grams.

GIVEN:

Amount of C6H6 = 42.1g

Amount of Br2 = 73.0g

TO FIND :

Theoretical yield of C6H5Br

SOLUTION:

We can simply solve the above problem as under:

According to the given equation:

C6H6 + Br2 --------> C6H5Br + HBr

This means :

1 mole of C6H6 on reaction with 1 more of Br2 gives 1 mole of C6H5Br and 1 mole of HBr.

It is given that 42.1g of C6H6 reacted with 73.0 grams of Br2.

Converting into moles we have :

42.1 grams of C6H6 = 42.1 /78.

(Molar mass of benzene is 78 g/mol )

42.1 grams of C6H6 = 0.54 mol.

73 grams of Br2 = 73/159.8

(Molar mass of bromine molecule is 158.9 g/mol)

73 grams of Br2 = 0.46 mol

This implies, that bromine is the limiting reagent in the given reaction.

According to stochiometry,

we know that 1 mole of Br2 will yield 1 mole of C6H5Br.

Theoretical yield = MOLES OF limiting reagent × Molar mass of product

Theoretical yield of C6H5Br = 0.46× 156.9

= 72.2

Hence, the Theoretical yield of C6H5Br is 72.2 grams

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