Question

DDT, an insecticide harmful to fish, birds, and humans, is produced by the following reaction:

$\mathbf{2}{\mathbf{\text{C}}}_{\mathbf{6}}{\mathbf{\text{H}}}_{\mathbf{5}}\mathbf{\text{Cl}}\mathbf{+}{\mathbf{\text{C}}}_{\mathbf{2}}{\mathbf{\text{HOCl}}}_{\mathbf{3}}\mathbf{\to }{\mathbf{\text{C}}}_{\mathbf{14}}{\mathbf{\text{H}}}_{\mathbf{9}}{\mathbf{\text{Cl}}}_{\mathbf{5}}\mathbf{+}{\mathbf{\text{H}}}_{\mathbf{2}}\mathbf{\text{O}}$

In a government lab, 1142 g of chlorobenzene is reacted with 485 g of chloral.
a. What mass of DDT is formed, assuming 100% yield?
b. Which reactant is limiting? Which is in excess?
c. What mass of the excess reactant is left over?
d. If the actual yield of DDT is 200.0 g, what is the percent yield?

Short answer

1. The mass of DDT formed is$\mathbf{1166}\mathbf{.}\mathbf{27}\mathbf{\text{g}}$.
2. Chloral $\left({\text{C}}_2{\text{HOCl}}_3\right)$is the limiting, and the excess is in chlorobenzene$\left({\text{C}}_6{\text{H}}_5\text{Cl}\right)$.
3. The mass of the excess reactantchlorobenzene is$\mathbf{401}\mathbf{.}\mathbf{83}\mathbf{\text{g}}$.
4. The percent yield is$\mathbf{17}\mathbf{.}\mathbf{15}\mathbf{\%}$ .

Step-by-step solution

Definition

The mass of any material expressed in atomic mass units is quantitatively equivalent to its molar mass in grams per mole.

Determination of mass of DDT Produced

a)

The given reaction:

$2{\text{C}}_6{\text{H}}_5\text{Cl}+{\text{C}}_2{\text{HOCl}}_3\to {\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5+{\text{H}}_2\text{O}$

$\mathbf{\text{Molesofchlorobenzene}}\mathbf{=}\frac{\mathbf{\text{Mass}}}{\mathbf{\text{molarmass}}}$

$=\frac{1147\text{g}}{11256{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

$=10.15\mathrm{moles}$

Moles of chloral$=\frac{48\text{g}}{147.38{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

$=3.29\mathrm{moles}$moles

Two moles of chlorobenzene react with one mole of chloral in the equation.

5.075 moles of chloral are required for 10.15 moles of chlorobenzene. As a result, chloral is the limiting reactant, whereas chlorobenzene is overproduced.3.29 moles of chloral react with 6.58 moles chlorobenzene to produce 3.29 moles DDT.

$\mathbf{\text{MolesofDDT}}\mathbf{=}\left(\text{Mass}\right)\mathbf{\times }\left(\text{molarmass}\right)$

$3.29\times 354.49\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$=1166.27g$

Therefore, the moles of DDT have a mass of$\mathbf{1166}\mathbf{.}\mathbf{27}\mathbf{\text{g}}$

Determination of limitant and excess

b)

$2{\text{C}}_6{\text{H}}_5\text{Cl}+{\text{C}}_2{\text{HOCl}}_3\to {\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5+{\text{H}}_2\text{O}$

$n\left(C_6{H}_{5}Cl\right)=\frac{m\left(C_6{H}_{5}Cl\right)}{Mr\left(C_6{H}_{5}Cl\right)}$

$\frac{1142\text{g}}{112.5{\displaystyle \raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.}}$

$=10.15\text{mol}$

$n\left(C_{2}HOC{l}_3\right)=\frac{m\left(C_{2}HOC{l}_3\right)}{Mr\left(C_{6}HOC{l}_3\right)}$

$=3.29\text{mol}$

Therefore, chloral$\left({\text{C}}_2{\text{HOCl}}_3\right)$is the limitant, and the excess is chlorobenzene$\left({\text{C}}_6{\text{H}}_5\text{Cl}\right)$.

Determination of mass of excess

c)

Here, $6.58$moles of chlorobenzene are consumed.

Excess moles of chlorobenzene $=10.15-6.58$

$=3.57\text{moles}$

Mass of chlorobenzene $=3.57\times 112.56\raisebox{1ex}{$\text{g}$}\!\left/ \!\raisebox{-1ex}{$\text{mol}$}\right.$

$=401.83\text{g}$

Therefore, the massof chlorobenzeneis$\mathbf{\text{401.83g}}$ .

Determination of Percentage yield 

d)

Actual yield$=200.0\mathrm{g}$

Theoretical yield $=1166.27\mathrm{g}$

$\mathbf{\text{Percentageyield}}\mathbf{=}\left(\frac{\text{Actualyield}}{\text{Theoriticalyield}}\right)\mathbf{\times }\mathbf{100}$

$=\left(\frac{200.0g}{1166.27g}\right)\times 100$

$=17.15\%$

Therefore, percentage yield is $\mathbf{17}\mathbf{.}\mathbf{15}\mathbf{\%}$.