Chapter 3, Problem 88E

(a)

Interpretation Introduction

Interpretation: The mass of acrylonitrile produced from $1.00\text{kg}$ of propylene, $\text{1}\text{.50}\text{kg}$ of ammonia and $\text{2}\text{.00}\text{kg}$ of oxygen in the given reaction with $100\%$ yield is to be calculated.

Concept introduction: The relation between moles of reaction species is estimated through the stoichiometric coefficients written with each species in the reaction. These coefficients are useful in calculating reaction’s yield. The stoichiometry calculations are useful in determining the limiting reactant on which the yield of reaction is actually dependent.

Answer to Problem 88E

The mass of acrylonitrile produced from $1.00\text{kg}$ of propylene, $\text{1}\text{.50}\text{kg}$ of ammonia and $\text{2}\text{.00}\text{kg}$ of oxygen in the given reaction with $100\%$ yield is $1261.9\text{g}$ .

Explanation of Solution

The given reaction is shown below.

  ${\text{2C}}_3{\text{H}}_6\left(\text{g}\right)+{\text{2NH}}_{\text{3}}\left(\text{g}\right)+{\text{3O}}_{\text{2}}\left(\text{g}\right)\to {\text{2C}}_3{\text{H}}_3\text{N}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

The values of molar masses of ${\text{NH}}_{\text{3}}\text{,}{\text{O}}_{\text{2}},{\text{C}}_3{\text{H}}_6,{\text{C}}_3{\text{H}}_3\text{N}$ and ${\text{H}}_{\text{2}}\text{O}$ are $17\text{g/mol,}\text{32}\text{g/mol},42\text{g/mol},53\text{g/mol}$ and $\text{18}\text{g/mol}$ respectively.

The formula to calculate moles is shown below.

  $\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}$  (1)

Substitute the value of mass and molar mass of ammonia in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{1.50\text{kg}}{17\text{g/mol}}\\ =\frac{\left(1.50\text{kg}\times \frac{\text{1000}\text{g}}{1\text{kg}}\right)}{17\text{g/mol}}\\ =88.23\text{mol}\end{array}$

Substitute the value of mass and molar mass of oxygen in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{2.00\text{kg}}{32\text{g/mol}}\\ =\frac{\left(2.00\text{kg}\times \frac{\text{1000}\text{g}}{1\text{kg}}\right)}{32\text{g/mol}}\\ =62.50\text{mol}\end{array}$

Substitute the value of mass and molar mass of propylene in equation (1).

  $\begin{array}{c}\text{Moles}=\frac{1.00\text{kg}}{42\text{g/mol}}\\ =\frac{\left(1.00\text{kg}\times \frac{\text{1000}\text{g}}{1\text{kg}}\right)}{42\text{g/mol}}\\ =23.81\text{mol}\end{array}$

For the reactants ammonia and ${\text{C}}_3{\text{H}}_6$ , the required mole ratio is calculated as follows.

  $\begin{array}{c}\frac{\text{mol}{\text{NH}}_{\text{3}}}{\text{mol}{\text{C}}_3{\text{H}}_6}=\frac{2}{2}\\ =1\end{array}$

For the reactants ammonia and ${\text{C}}_3{\text{H}}_6$ , the actual mole ratio is calculated as follows.

  $\begin{array}{c}\frac{\text{mol}{\text{NH}}_{\text{3}}}{\text{mol}{\text{C}}_3{\text{H}}_6}=\frac{88.23\text{mol}}{23.81\text{mol}}\\ =3.71\end{array}$

Since the actual ratio is more than the required ratio, therefore, ammonia is in excess as compared to ${\text{C}}_3{\text{H}}_6$ .

For the reactants ${\text{C}}_3{\text{H}}_6$ and oxygen, the required mole ratio is calculated as,

  $\begin{array}{c}\frac{\text{mol}{\text{O}}_2}{\text{mol}{\text{C}}_3{\text{H}}_6}=\frac{3}{2}\\ =1.5\end{array}$

For the reactants ${\text{C}}_3{\text{H}}_6$ and oxygen, the actual mole ratio is calculated as,

  $\begin{array}{c}\frac{\text{mol}{\text{O}}_2}{\text{mol}{\text{C}}_3{\text{H}}_6}=\frac{62.50\text{mol}}{23.81\text{mol}}\\ =2.62\end{array}$

Since the actual ratio is more than the required ratio, therefore, oxygen is also in excess as compared to ${\text{C}}_3{\text{H}}_6$ .

Thus, the limiting reagent is ${\text{C}}_3{\text{H}}_6$ .

From the reaction, it is clear that $\text{2}\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ give $\text{2}\text{mol}$ of ${\text{C}}_3{\text{H}}_3\text{N}$ .

So, $23.81\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ would give $\frac{\text{2}\text{mol}}{2\text{mol}}\times 23.81\text{mol}$ that is $23.81\text{mol}$ of ${\text{C}}_3{\text{H}}_3\text{N}$ .

The mass of ${\text{C}}_3{\text{H}}_3\text{N}$ is calculated using equation (1) as follows.

  $\begin{array}{c}\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 23.81\text{mol}=\frac{\text{Mass}}{53\text{g/mol}}\\ \text{Mass}=23.81\text{mol}\times 53\text{g/mol}\\ =1261.9\text{g}\end{array}$

Therefore, the mass of acrylonitrile is $1261.9\text{g}$ .

(b)

Interpretation Introduction

Interpretation: The mass of water produced from $1.00\text{kg}$ of propylene, $\text{1}\text{.50}\text{kg}$ of ammonia and $\text{2}\text{.00}\text{kg}$ of oxygen in the given reaction with $100\%$ yield is to be calculated and the reactant with its mass in excess is to be stated.

Concept introduction: The relation between moles of reaction species is estimated through the stoichiometric coefficients written with each species in the reaction. These coefficients are useful in calculating reaction’s yield. The stoichiometry calculations are useful in determining the limiting reactant on which the yield of reaction is actually dependent.

Answer to Problem 88E

The mass of water produced from $1.00\text{kg}$ of propylene, $\text{1}\text{.50}\text{kg}$ of ammonia and $\text{2}\text{.00}\text{kg}$ of oxygen in the given reaction with $100\%$ yield is $1285.7\text{g}$ . The excess reactants are ammonia and oxygen with left masses $1095.1\text{g}$ and $857.28\text{g}$ respectively.

Explanation of Solution

The given reaction is shown below.

  ${\text{2C}}_3{\text{H}}_6\left(\text{g}\right)+{\text{2NH}}_{\text{3}}\left(\text{g}\right)+{\text{3O}}_{\text{2}}\left(\text{g}\right)\to {\text{2C}}_3{\text{H}}_3\text{N}\left(\text{g}\right)+{\text{6H}}_{\text{2}}\text{O}\left(\text{g}\right)$

From the reaction, it is clear that $\text{2}\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ give $\text{6}\text{mol}$ of ${\text{H}}_2\text{O}$ .

So, $23.81\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ would give $\frac{\text{6}\text{mol}}{2\text{mol}}\times 23.81\text{mol}$ that is $71.43\text{mol}$ of ${\text{H}}_2\text{O}$ .

The mass of ${\text{H}}_2\text{O}$ is calculated using equation (1) as follows.

  $\begin{array}{c}\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 71.43\text{mol}=\frac{\text{Mass}}{18\text{g/mol}}\\ \text{Mass}=71.43\text{mol}\times 18\text{g/mol}\\ =1285.7\text{g}\end{array}$

Therefore, the mass of water produced is $1285.7\text{g}$ .

In the reaction, $\text{2}\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ reacts with $\text{2}\text{mol}$ of ammonia.

So, $23.81\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ would react with $\frac{\text{2}\text{mol}}{2\text{mol}}\times 23.81\text{mol}$ that is $23.81\text{mol}$ of ammonia. The moles of ammonia present are $88.23\text{mol}$ . So, the moles of ammonia in excess are $\left(88.23\text{mol}-23.81\text{mol}\right)=64.42\text{mol}$ .

The mass of excess ammonia is calculated using equation (1) as follows.

  $\begin{array}{c}\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 64.42\text{mol}=\frac{\text{Mass}}{17\text{g/mol}}\\ \text{Mass}=64.42\text{mol}\times 17\text{g/mol}\\ =1095.1\text{g}\end{array}$

Therefore, the mass of ammonia in excess is $1095.1\text{g}$ .

In the reaction, $\text{2}\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ reacts with $\text{3}\text{mol}$ of oxygen.

So, $23.81\text{mol}$ of ${\text{C}}_3{\text{H}}_6$ would react with $\frac{\text{3}\text{mol}}{2\text{mol}}\times 23.81\text{mol}$ that is $35.71\text{mol}$ of oxygen. The moles of oxygen present are $62.50\text{mol}$ . So, the moles of oxygen in excess are $\left(62.50\text{mol}-35.71\text{mol}\right)=26.79\text{mol}$ .

The mass of excess oxygen is calculated using equation (1) as follows.

  $\begin{array}{c}\text{Moles}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ 26.79\text{mol}=\frac{\text{Mass}}{32\text{g/mol}}\\ \text{Mass}=26.79\text{mol}\times 32\text{g/mol}\\ =857.28\text{g}\end{array}$

Therefore, the mass of oxygen in excess is $857.28\text{g}$ .