Need help before April 10th! Please and thank you!

Question

___C6H6+ ___Cl2→ ___C6H5Cl + ___HCl

When 39.0 g of C6H6reacts with an excess of Cl2, the actual yield of C6H5Cl is 42.0 g. What is the percent yield of C6H5Cl?

Answer 1

Hi Chloe N.

Since 39 g of Benzene is reacted with an excess of Chlorine gas, benzene is your limiting reagent. The benzene will be consumed in the reaction to give chloro-benzene.

Your balanced reaction shows that 1 mole of Benzene reacts with 1 mole of Chlorine gas to produce 1 mole of Chloro-benzene and 1 mole of Hydrogen Chloride.

Percent Yield = Actual yield * 100/Theoretical Yield

Actual Yield = 42.0 g

Theoretical Yield = the amount expected from 100% reaction completion so we need to see what that is

39g of Benzene is

39g * 1 mole/78.11g = 0.499 basically half a mole

From your balanced reaction we know that 1 mole of benzene gives 1 mole of Chloro-benzene

So half a mole of benzene would theoretically yield half a mole of chloro-benzene if your reaction goes to 100% completion and a half mole of chloro-benzene is

0.499 mole x 112.56g/mole = 56.17 g

56.17 g is your theoretical yield

Percent yield = Actual yield*100/Theoretical yield

Percent yield = 42g*100/56.17g = 74.77%

I hope this helps.