Question

What is the theoretical yield of C6H5Br if 42.1 g of C6H6 react with 73.0 g of Br2?

Answer 1

1) Chemical reaction

C6H6 + Br2 ---> C6H5Br + HBr


2) Molar raios

1 mol C6H6 : 1 mol Br2 : 1 mol C6H5Br


3) Convert the data to moles

42.1 g of C6H6

molar mass of C6H6 = 6*12g/mol + 6*1g/mol = 78 g/mol

42.1 g / 78 g/mol = 0.54 mol


73.0 g of Br2


molar mass of Br2 = 2 * 79.9 g/mol = 159.8 g/mol

73.0 g / 159.8 g/mol = 0.46 mol of Br2


=> limiting reagent is the Br2.


4) Product

1 mol of Br2 (limiting reagent) yields 1 mole of C6H5Br, then you will obtain 0.46 mol of C6H5Br


5) Convert the product to grams

molar mass of C6H5Br = 6*12g/mol + 5*1g/mol + 79.9 g/mol = 156.9 g/mol

0.46mol*156.9g/mol = 72.2 g.


Answer: 72.2 g

 

Answer 2

The theoretical yield of $C_6H_5Br$ if 42.1 g of  $C_6H_6$ reacting with 73.0 g of $Br_2$ is 72.2 grams.

What is benzene?

Benzene is a chemical compound that is light yellow.

The chemical formula $C_6H_6$.

Given,

Chemical reaction

$C_6H_6 + Br_2 = C_6H_5Br + HBr$

The  Molar ratios are

1 mol $C_6H_6$ : 1 mol $Br_2$ : 1 mol $C_6H_5Br$

Step1: Calculating the moles

42.1 g of $C_6H_6$

molar mass of $C_6H_6$ = $6 \times12g/mol + 6 \times 1g/mol = 78 g/mol$

$\dfrac{42.1 g}{78 g/mol}  = 0.54 mol$

73.0 g of $Br_2$

molar mass of  $br_2$ = $2 \times 79.9 g/mol = 159.8 g/mol

[tex]\dfrac{73.0 g}{159.8 g/mol} = 0.46 mol\; of\; Br_2$

The limiting reagent is $Br_2$

Step 2: If 1 mol of  $Br_2$(limiting reagent) yields 1 mole of  $C_6H_5Br$, then you will obtain 0.46 mol of $C_6H_5Br$

Step 3: Convert the product into grams

molar mass of $C_6H_5Br$ =  $6 \times12g/mol + 5 \times 1g/mol + 79.9 g/mol = 156.9 g/mol$

$0.46\;mol \times156.9\;g/mol = 72.2\; g$

Thus, the theoretical yield is 72.2 grams.

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