Chapter 5, Problem 124E

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

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?

Interpretation Introduction

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The mass of DDT formed in the given reaction.

Explanation of Solution

Given

The balanced chemical equation is,

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

The given mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $1142\text{g}$.

The given mass of ${\text{C}}_2{\text{HOCl}}_3$ is $485\text{g}$.

The molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ $\begin{array}{l}=6\text{C}+5\text{H}+\text{Cl}\\ =\left(\left(6\times 12\right)+\left(5\times 1\right)+35.5\right)\text{g}/\text{mol}\\ =112.5\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{C}}_2{\text{HOCl}}_3$ $\begin{array}{l}=2\text{C}+\text{H}+\text{O}+3\text{Cl}\\ =\left(\left(2\times 12\right)+1+16+\left(3\times 35.5\right)\right)\text{g}/\text{mol}\\ =147.5\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

Substitute the value of the given mass and the molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ and ${\text{C}}_2{\text{HOCl}}_3$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}=\frac{1142\text{g}}{112.5\text{g}/\text{mol}}\\ =10.15\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{C}}_2{\text{HOCl}}_3=\frac{485\text{g}}{147.5\text{g}/\text{mol}}\\ =3.29\text{mol}\end{array}$

According to the stated reaction,

$2\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ react with $1\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$.

$1\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacts with ${\text{C}}_2{\text{HOCl}}_3$ $=\left(\frac{1}{2}\right)\text{mol}$

$10.15\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacts with ${\text{C}}_2{\text{HOCl}}_3$ $\begin{array}{l}=\left(\frac{1\times 10.15}{2}\right)\text{mol}\\ =5.075\text{mol}\end{array}$

The number of moles of ${\text{C}}_2{\text{HOCl}}_3$ present are $3.29\text{mol}$.

Therefore, this is the limiting reactant in the reaction.

According to the stated reaction,

$1\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ produces $1\text{mol}$ of ${\text{C}}_4{\text{H}}_{19}{\text{Cl}}_5$.

$3.29\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ produces $3.29\text{mol}$ of ${\text{C}}_4{\text{H}}_{19}{\text{Cl}}_5$.

The molar mass of ${\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5$ $\begin{array}{l}=14\text{C}+9\text{H}+5\text{Cl}\\ =\left(\left(14\times 12\right)+\left(9\times 1\right)+\left(5\times 35.5\right)\right)\text{g}/\text{mol}\\ =354.5\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of ${\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5$ and the molar mass of ${\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}{\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5=\left(3.29\text{mol}\right)\times \left(354.5\text{g}/\text{mol}\right)\\ =\underset{\_}{1.17\times 10^{3}g}\end{array}$

Conclusion

The mass of DDT formed in the given chemical reaction is $\underset{\_}{1.17\times 10^{3}g}$.

Interpretation Introduction

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The limiting and the excess reactant.

Explanation of Solution

Given

The balanced chemical equation is,

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

The given mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $1142\text{g}$.

The given mass of ${\text{C}}_2{\text{HOCl}}_3$ is $485\text{g}$.

The molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ $\begin{array}{l}=6\text{C}+5\text{H}+\text{Cl}\\ =\left(\left(6\times 12\right)+\left(5\times 1\right)+35.5\right)\text{g}/\text{mol}\\ =112.5\text{g}/\text{mol}\end{array}$

The molar mass of ${\text{C}}_2{\text{HOCl}}_3$ $\begin{array}{l}=2\text{C}+\text{H}+\text{O}+3\text{Cl}\\ =\left(\left(2\times 12\right)+1+16+\left(3\times 35.5\right)\right)\text{g}/\text{mol}\\ =147.5\text{g}/\text{mol}\end{array}$

Formula

The number of moles of a substance is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}\text{of}\text{the}\text{substance}}$

Substitute the value of the given mass and the molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ and ${\text{C}}_2{\text{HOCl}}_3$ in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}=\frac{1142\text{g}}{112.5\text{g}/\text{mol}}\\ =10.15\text{mol}\end{array}$

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{C}}_2{\text{HOCl}}_3=\frac{485\text{g}}{147.5\text{g}/\text{mol}}\\ =3.29\text{mol}\end{array}$

According to the stated reaction,

$2\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ react with $1\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$.

$1\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacts with ${\text{C}}_2{\text{HOCl}}_3$ $=\left(\frac{1}{2}\right)\text{mol}$

$10.15\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacts with ${\text{C}}_2{\text{HOCl}}_3$ $\begin{array}{l}=\left(\frac{1\times 10.15}{2}\right)\text{mol}\\ =5.075\text{mol}\end{array}$

The number of moles of ${\text{C}}_2{\text{HOCl}}_3$ present are $3.29\text{mol}$.

Therefore, this is the limiting reactant in the reaction and ${\text{C}}_6{\text{H}}_5\text{Cl}$ is the excess reactant.

Conclusion

The limiting reactant is ${\text{C}}_2{\text{HOCl}}_3$ and the excess reactant is ${\text{C}}_6{\text{H}}_5\text{Cl}$.

Interpretation Introduction

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The mass of the excess reactant left over.

Explanation of Solution

Given

The balanced chemical equation is,

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

According to the stated reaction,

$1\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ reacts with $2\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$.

$\text{3}\text{.29}\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ reacts with ${\text{C}}_6{\text{H}}_5\text{Cl}$ $\begin{array}{l}=\left(2\times \text{3}\text{.29}\right)\text{mol}\\ =6.58\text{mol}\end{array}$

The moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ given are $10.15\text{mol}$.

Therefore, some amount of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is left unreacted. It is the excess reactant.

The moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ left unreacted are calculated by the formula,

$\text{Unreacted}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}=\text{Given}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}-\text{Reacted}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}$

Substitute the values of the given moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ and the reacted moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in the above expression.

$\begin{array}{c}\text{Unreacted}\text{moles}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}=\left(\text{10}\text{.15}-\text{6}\text{.58}\right)\text{mol}\\ =\text{3}\text{.57}\text{mol}\end{array}$

The molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ $\begin{array}{l}=6\text{C}+5\text{H}+\text{Cl}\\ =\left(\left(6\times 12\right)+\left(5\times 1\right)+35.5\right)\text{g}/\text{mol}\\ =112.5\text{g}/\text{mol}\end{array}$

The mass of a substance is calculated by the formula,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

Substitute the value of the number of moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ and the molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in the above expression.

$\begin{array}{c}\text{Mass}\text{of}{\text{C}}_6{\text{H}}_5\text{Cl}=\left(\text{3}\text{.57}\text{mol}\right)\times \left(112.5\text{g}/\text{mol}\right)\\ =\underset{\_}{401.6g}\end{array}$

Conclusion

The mass of the excess reactant, that is ${\text{C}}_6{\text{H}}_5\text{Cl}$ , left unreacted is $\underset{\_}{401.6g}$.

Interpretation Introduction

Concept introduction: The mass of a substance can be obtained by using the number of moles of the substance present and its molar mass. The formula used to calculate the mass of a given substance is,

$\text{Mass}\text{of}\text{the}\text{substance}=\left(\text{Number}\text{of}\text{moles}\right)\times \left(\text{Molar}\text{mass}\text{of}\text{the}\text{substance}\right)$

To determine: The percent yield of DDT $\left({\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5\right)$.

Explanation of Solution

Given

The balanced chemical equation for the given chemical reaction is,

${\text{C}}_2{\text{H}}_6\left(g\right)+{\text{Cl}}_2\left(g\right)\to {\text{C}}_2{\text{H}}_5\text{Cl}\left(g\right)+\text{HCl}\left(g\right)$

The actual yield of DDT is given to be $200\text{g}$.

The mass of DDT formed in the given chemical reaction is calculated to be $\underset{\_}{1.17\times 10^{3}g}$.

This is the theoretical yield.

Formula

The percent yield of ${\text{C}}_2{\text{H}}_5\text{Cl}$ is calculated by the formula,

$\text{Percent}\text{yield}=\frac{\text{Actual}\text{yield}}{\text{Theoretical}\text{yield}}\times 100$

Substitute the value of the actual yield and the theoretical yield of DDT in the above expression.

$\begin{array}{c}\text{Percent}\text{yield}=\left(\frac{\text{200}\text{g}}{1.17\times {10}^3\text{g}}\times 100\right)\%\\ =\underset{\_}{17.1\%}\end{array}$

Conclusion

The percent yield of DDT is $\underset{\_}{17.1\%}$.