Chapter 3, Problem 86E

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

$\begin{array}{l}{\text{2C}}_{\text{6}}{\text{H}}_{\text{5}}\text{Cl}\text{+}{\text{C}}_{\text{2}}{\text{HOCl}}_{\text{3}}\to {\text{C}}_{\text{14}}{\text{H}}_{\text{9}}{\text{Cl}}_{\text{5}}{\text{+H}}_{\text{2}}\text{O}\\ \text{Chlorobenzene }\text{Chloral }\text{DDT}\end{array}$

In a government lab, 1142 g of chlorobenzene is reactedwith 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 thepercent yield?

Interpretation Introduction

Interpretation: The mass of DDT formed, assuming $100\%$ yield is to be calculated.

Concept introduction: The limiting reactant is the reactant which depleted completely in a reaction and the amount of the product formed is also depends upon this. The excess reactant is the reactant which does not reacted completely in a reaction and some of the amount of such reactant left unreacted.

Answer to Problem 86E

The mass of DDT formed, assuming $100\%$ yield is $1166.30\text{g}$ .

Explanation of Solution

The given balanced chemical equation is given below.

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

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

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

The standard molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $112.5\text{g}/\text{mol}$ .

The standard molar mass of ${\text{C}}_{14}{\text{H}}_9{\text{Cl}}_5$ is $354.5\text{g}/\text{mol}$ .

The standard molar mass of ${\text{C}}_2{\text{HOCl}}_3$ is $147.5\text{}\text{g}/\text{mol}$ .

The formula to compute number of moles is as follows.

  $\text{Moles}=\frac{\text{Given mass}\left(\text{g}\right)}{\text{Molar mass}\left(\text{g}/\text{mol}\right)}$  (1)

Substitute the values of given mass and molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{1142\left(\text{g}\right)}{112.5\left(\text{g}/\text{mol}\right)}\\ =10.15\text{mol}\end{array}$

Substitute the values of given mass and molar mass of ${\text{C}}_2{\text{HOCl}}_3$ in equation(1) as shown below.

  $\begin{array}{c}\text{Moles }=\frac{485\left(\text{g}\right)}{147.5\left(\text{g}/\text{mol}\right)}\\ =3.29\text{mol}\end{array}$

From the balanced equation, it is clear that $2\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is reacted with $\text{1}\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ .

So, $10.15\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacted with ${\text{C}}_2{\text{HOCl}}_3$ equals to $\frac{1}{2}\times 10.15=5.07\text{moles}$ .

But, only $3.29\text{mol}$ of chloral is present.

So, ${\text{C}}_2{\text{HOCl}}_3$ present in limited amount whereas, ${\text{C}}_6{\text{H}}_5\text{Cl}$ present in excess amount.

Thus, amount of product formed depends on amount of chloral.

One mole of chloral produces one mole of DDT.

So, $3.29\text{mol}$ of chloral produce DDT equals to $\frac{1}{1}\times 3.29=3.29\text{mol}$ .

The formula to calculate the mass of a substance is as follows.

  $\text{Mass}\left(\text{g}\right)=\text{Molar mass}\left(\text{g}/\text{mol}\right)\times \text{Moles}$  (2)

Substitute the molar mass and moles of DDT in equation (2).

  $\begin{array}{c}\text{Mass}=\text{354}\text{.5}\left(\text{g}/\text{mol}\right)\times 3.29\left(\text{mol}\right)\\ =1166.30\text{g}\end{array}$

Therefore, the mass of DDT formed, assuming $100\%$ yield is $1166.30\text{g}$ .

Interpretation Introduction

Interpretation: The limiting reactant and excess reactant are to be predicted.

Concept introduction: The limiting reactant is the reactant which depleted completely in a reaction and the amount of the product formed is also depends upon this. The excess reactant is the reactant which does not reacted completely in a reaction and some of the amount of such reactant left unreacted.

Answer to Problem 86E

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

Explanation of Solution

The given balanced chemical equation is given below.

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

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

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

The standard molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $112.5\text{g}/\text{mol}$ .

The standard molar mass of ${\text{C}}_2{\text{HOCl}}_3$ is $147.5\text{}\text{g}/\text{mol}$ .

The formula to compute number of moles is as follows.

  $\text{Moles}=\frac{\text{Given mass}\left(\text{g}\right)}{\text{Molar mass}\left(\text{g}/\text{mol}\right)}$  (1)

Substitute the values of given mass and molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{1142\left(\text{g}\right)}{112.5\left(\text{g}/\text{mol}\right)}\\ =10.15\text{mol}\end{array}$

So, moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $10.15\text{mol}$ .

Substitute the value of given mass and molar mass of ${\text{C}}_2{\text{HOCl}}_3$ in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{485\left(\text{g}\right)}{147.5\left(\text{g}/\text{mol}\right)}\\ =3.29\text{mol}\end{array}$

From the balanced equation, it is clear that $2\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is reacted with $1\text{}\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ .

So, $10.15\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ reacted with ${\text{C}}_2{\text{HOCl}}_3$ equals to $\frac{1}{2}\times 10.15=5.07\text{mol}$ .

But, only $3.29\text{mol}$ of chloral is present.

So, ${\text{C}}_2{\text{HOCl}}_3$ is present in limited amount whereas, ${\text{C}}_6{\text{H}}_5\text{Cl}$ is present in excess amount.

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

Interpretation Introduction

Interpretation: The mass of excess reactant that is left is to be calculated.

Concept introduction: The limiting reactant is the reactant which depleted completely in a reaction and the amount of the product formed is also depends upon this. The excess reactant is the reactant which does not reacted completely in a reaction and some of the amount of such reactant left unreacted.

Answer to Problem 86E

The mass of excess reactant left is $401.625\text{g}$ .

Explanation of Solution

The given balanced chemical equation is given below.

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

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

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

The standard molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is $112.5\text{g}/\text{mol}$ .

The standard molar mass of ${\text{C}}_2{\text{HOCl}}_3$ is $147.5\text{}\text{g}/\text{mol}$ .

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

The formula to compute number of moles is as follows.

  $\text{Moles}=\frac{\text{Given mass}\left(\text{g}\right)}{\text{Molar mass}\left(\text{g}/\text{mol}\right)}$  (1)

Substitute the value of given mass and molar mass of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in equation(1).

  $\begin{array}{c}\text{Moles}=\frac{1142\left(\text{g}\right)}{112.5\left(\text{g}/\text{mol}\right)}\\ =10.15\text{mol}\end{array}$

Substitute the value of given mass and molar mass of ${\text{C}}_2{\text{HOCl}}_3$ in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{485\left(\text{g}\right)}{147.5\left(\text{g}/\text{mol}\right)}\\ =3.29\text{mol}\end{array}$

From the balanced equation, it is clear that $1\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ is reacted with $2\text{}\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ .

So, the $3.29\text{mol}$ of ${\text{C}}_2{\text{HOCl}}_3$ reacted with ${\text{C}}_6{\text{H}}_5\text{Cl}$ equals to $2\times 3.29=6.58\text{mol}$ .

But, only $10.15\text{mol}$ of ${\text{C}}_6{\text{H}}_5\text{Cl}$ is present.

So, the reactant present in excess amount is ${\text{C}}_6{\text{H}}_5\text{Cl}$ .

The formula to calculate the left amount of excess reactant using number of moles is as follows.

  $\text{Left mass}=\left(\text{Initial moles}-\text{Reacted moles}\right)\times \text{Molar mass}\left(\text{g}/\text{mol}\right)$ (3)

Substitute the initial moles and reacted moles of ${\text{C}}_6{\text{H}}_5\text{Cl}$ in equation (3).

  $\begin{array}{c}\text{Left mass}=\left(10.15-6.58\right)\text{mol}\times 112.5\left(\text{g}/\text{mol}\right)\\ =3.57\text{mol}\times 112.5\left(\text{g}/\text{mol}\right)\\ =401.625\text{g}\end{array}$

Therefore, the mass of excess reactant left is $401.625\text{g}$ .

Interpretation Introduction

Interpretation: The percent yield of DDT is to be calculated.

Concept introduction: The percent yield of a substance is calculated by using actual yield and theoretical yield. The formula to calculate the percent yield is shown below.

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

Answer to Problem 86E

The percent yield of DDT is $17.15\%$ .

Explanation of Solution

The given balanced chemical equation is given below.

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

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

The calculated or theoretical yield of DDT is $1166.30\text{g}$ .

The formula to calculate the percent yield is as follows.

  $\text{Percent yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100$  (4)

Substitute the values of actual yield and theoretical yield of DDT in equation (4).

  $\begin{array}{c}\text{Percent yield}=\frac{200\text{g}}{1166.30\text{g}}\times 100\\ =17.15\%\end{array}$

Therefore, the percent yield of DDT is $17.15\%$ .