Question

ddt, an insecticide harmful to fish, birds, and humans, is produced by the following reaction: 2c₆h₅cl + c₂hocl₃ → c₁₄h₉cl₅ + h₂o chlorobenzene chloral ddt in a government lab, 1185 g of chlorobenzene is reacted with 489 g of chloral. a. what mass of ddt is formed, assuming 100% yield? mass = 1176.12 g b. which reactant is limiting? which is in excess? c₂hocl₃ is limiting. 2c₆h₅cl is in excess. c. what mass of the excess reactant is left over? mass = 438.11 g in excess d. if the actual yield of ddt is 201.0 g, what is the percent yield? percent yield = 17.09 %

Explanation:

Step1: Calculate molar masses

Molar mass of $C_6H_5Cl$: $(12\times6 + 1\times5+35.5)=112.5$ g/mol. Molar mass of $C_2HOCl_3$: $(12\times2 + 1+16 + 35.5\times3)=147.5$ g/mol. Molar mass of $C_{14}H_9Cl_5$: $(12\times14+1\times9 + 35.5\times5)=354.5$ g/mol.

Step2: Calculate moles of reactants

Moles of $C_6H_5Cl=\frac{1185}{112.5}=10.533$ mol. Moles of $C_2HOCl_3=\frac{489}{147.5}=3.315$ mol.

Step3: Determine limiting reactant

From the balanced - equation, the mole ratio of $C_6H_5Cl$ to $C_2HOCl_3$ is 2:1. For 3.315 mol of $C_2HOCl_3$, the moles of $C_6H_5Cl$ required is $2\times3.315 = 6.63$ mol. Since we have 10.533 mol of $C_6H_5Cl$, $C_2HOCl_3$ is the limiting reactant and $C_6H_5Cl$ is in excess.

Step4: Calculate mass of DDT formed

From the balanced - equation, 1 mole of $C_2HOCl_3$ forms 1 mole of $C_{14}H_9Cl_5$. Moles of $C_{14}H_9Cl_5$ formed = 3.315 mol. Mass of $C_{14}H_9Cl_5$ formed = $3.315\times354.5 = 1176.12$ g.

Step5: Calculate mass of excess reactant left

Moles of $C_6H_5Cl$ reacted = 6.63 mol. Moles of $C_6H_5Cl$ left = $10.533 - 6.63=3.903$ mol. Mass of $C_6H_5Cl$ left = $3.903\times112.5 = 438.11$ g.

Step6: Calculate percent yield

Percent yield=$\frac{201.0}{1176.12}\times100\% = 17.09\%$

Answer:

a. 1176.12 g
b. $C_2HOCl_3$ is limiting; $2C_6H_5Cl$ is in excess
c. 438.11 g
d. 17.09%