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?

53.4 % is the percent yield

Explanation:

This is the reaction:

C₆H₆ + Cl₂ → C₆H₅Cl + HCl

First of all we need to know the moles of benzene we used

39 g. 1 mol / 78 g = 0.5 moles

Ratio is 1:1 so 1 mol of benzene produces 1 mol of chloride

0.5 moles of chloride were produced by 0.5 moles of benzene

We must calculate the mass of chloride we produced

0.5 mol. 112.45 g / 1 mol = 56.2g

Let's calculate the percent yield

(Yield produced / Theoretical yield). 100

(30 g / 56.2 g). 100 = 53.4 %