Question

If 39.0 g of C6H6 reacts with excess chlorine and produces 30.0 g of C6H5Cl in the reaction C6H6 + Cl2 → C6H5Cl + HCl , what is the percent yield of C6H5Cl? 1. 50.0% 2. 53.4% 3. 69.4% 4. 13.2% 5. 76.9%

Answer 1

Answer:

The answer to your question is number 2.

Explanation:

Reaction

                                C₆H₆  +   Cl₂   ⇒   C₆H₅Cl   +  HCl

Molecular mass

C₆H₆ = (12 x 6) + (6 x 1) = 72 + 6 = 78 g

C₆H₅Cl = (12 x 6) + (5 x 1) + (1 x 35.5) = 112.5 g

Proportion

                           78 g of C₆H₆  -------------------  112.5 g of C₆H₅Cl

                           39 g of C₆H₆  ------------------    x

                           x = (39 x 112.5) / 78

                           x = 56.25 g of C₆H₅Cl

Percent yield = $\frac{experimental yield}{theoretical yield} x 100$

Percent yield = $\frac{30}{56.25} x 100$

Percent yield = 53.33 %

             

           

Answer 2

Answer:

The percent yield of chloro benzene is 53.4%.

Explanation:

$C_6H_6 + Cl_2\rightarrow C_6H_5Cl + HCl$

Mole of benzene = $\frac{39.0 g}{78 g/mol}=0.5 mol$

According to reaction, 1 mole benzene gives 1 mole of chloro benzene.

Then 0.5 moles of benzne will give:

$\frac{1}{1}\times 0.5 mol=0.5$ of chloro benzene.

Mass of 0.5 moles of chloro benzene = 0.5 mol × 112.5 g/mol = 56.25 g

Theoretical yield of of chloro benzene = 56.25 g

Experimental yield of of chloro benzene = 39.0 g

The percent yield of chloro benzene :

$Yield(\%)=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{30.0 g}{56.25 g}\times 100= 53.33\% \approx 53.4\%$