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

The answer to your question is number 2.

Step-by-step 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 =
`(experimental yield)/(theoretical yield) x 100`

Percent yield =
`(30)/(56.25) x 100`

Percent yield = 53.33 %

Answer 2

The percent yield of chloro benzene is 53.4%.

Step-by-step explanation:

`C_6H_6 + Cl_2\rightarrow C_6H_5Cl + HCl`

Mole of benzene =
`(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:

`(1)/(1)* 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}}* 100`

`=(30.0 g)/(56.25 g)* 100= 53.33\% \approx 53.4\%`