Chapter F, Problem M.11E

(a)

Interpretation Introduction

Interpretation:

Limiting reactant in formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ has to be identified.

Concept Introduction:

In a chemical reaction, the limiting reactant is the one that is present in lesser quantity than the required stoichiometric ratio between the various reactants that is present in it.  This limiting reactant decides the maximum yield of the product that will be formed in a chemical reaction.

Answer to Problem M.11E

Limiting reactant is identified as oxygen.

Explanation of Solution

First reaction in the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is given as follows:

    ${\text{P}}_{\text{4}}\text{(s)}+3{\text{O}}_{\text{2}}\text{(g)}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{(s)}$

From the balanced chemical equation it is found that $\text{1}\text{mol}$ of ${\text{P}}_{\text{4}}$ requires $\text{3}\text{mol}$ of ${\text{O}}_{\text{2}}$ to produce one mole of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$.  The number of moles of ${\text{O}}_{\text{2}}$ is calculated as follows;

    $\begin{array}{c}\text{No}\text{of}\text{moles}\text{of}{\text{O}}_{\text{2}}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{5.77\text{g}}{32\text{g}\cdot {\text{mol}}^{-1}}\\ =0.180\text{mol}\end{array}$

Number of moles of ${\text{P}}_{\text{4}}$ can be calculated as follows;

    $\begin{array}{c}\text{No}\text{of}\text{moles}\text{of}{\text{P}}_{\text{4}}=\frac{\text{Mass}}{\text{Molar}\text{mass}}\\ =\frac{5.77\text{g}}{124\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0465\text{mol}\end{array}$

Amount of oxygen required to react with phosphorus is calculated as follows:

    $\begin{array}{c}{n}_{{\text{O}}_{\text{2}}}=\text{0}\text{.0465}\text{mol}{\text{P}}_{\text{4}}\times \frac{\text{3}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}}\\ =0.140\text{mol}\end{array}$

It is found that the moles of oxygen that is supplied is more than the required for the phosphorus to react.  Hence, oxygen is the reagent that is present in excess while ${\text{P}}_{\text{4}}$ is the limiting reactant.

As ${\text{P}}_{\text{4}}$ is the limiting reactant, the moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ that can be formed is given as shown below:

    $\begin{array}{c}{n}_{{\text{(P}}_{\text{4}}{\text{O}}_{\text{6}}\text{)}}=\text{0}\text{.0465}\text{mol}{\text{P}}_{\text{4}}\times \frac{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}}\\ =0.0465\text{mol}\end{array}$

Therefore, the amount of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ formed is $\text{0}\text{.0465}\text{mol}$.

Second step in the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is given as follows:

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{(s)}+2{\text{O}}_{\text{2}}\text{(g)}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{(s)}$

The total moles of oxygen taken for the reaction is $\text{0}\text{.180}\text{mol}$.  In the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$, the moles of oxygen that is reacted is $\text{0}\text{.140}\text{mol}$.  Therefore, the amount of oxygen that remains is calculated as follows:

    $\begin{array}{c}\text{Amount}\text{of}{\text{O}}_{\text{2}}\text{remaining}=0.180\text{mol}-0.140\text{mol}\\ =0.040\text{mol}\end{array}$

The amount of oxygen that is required to react with $\text{0}\text{.0465}\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is calculated as follows:

    $\begin{array}{c}{n}_{{\text{O}}_{\text{2}}}=\text{0}\text{.0465}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\times \frac{\text{2}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\\ =0.0930\text{mol}\end{array}$

As the remaining amount of oxygen is lesser than the amount of oxygen that is required for the reaction to complete, the limiting reactant is found to be oxygen.

(b)

Interpretation Introduction

Interpretation:

Mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ that is produced has to be calculated.

Answer to Problem M.11E

Mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ obtained is $5.68\text{g}$.

Explanation of Solution

Reaction for the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is given as follows:

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{(s)}+2{\text{O}}_{\text{2}}\text{(g)}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{(s)}$

The total moles of oxygen taken for the reaction is $\text{0}\text{.180}\text{mol}$.  In the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$, the moles of oxygen that is reacted is $\text{0}\text{.140}\text{mol}$.  Therefore, the amount of oxygen that remains is calculated as follows:

    $\begin{array}{c}\text{Amount}\text{of}{\text{O}}_{\text{2}}\text{remaining}=0.180\text{mol}-0.140\text{mol}\\ =0.040\text{mol}\end{array}$

The mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ that can be formed is calculated as follows:

    $\begin{array}{c}{m}_{{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}=\text{0}\text{.040}\text{mol}{\text{O}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{10}}}{\text{2}\text{mol}{\text{O}}_{\text{2}}}\times 284\text{g}\cdot {\text{mol}}^{-1}\\ =5.68\text{g}\end{array}$

Thus the mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ obtained is $5.68\text{g}$.

(c)

Interpretation Introduction

Interpretation:

Mass of excess reactant that remains in the reaction has to be calculated.

Answer to Problem M.11E

Mass of unreacted excess reactant is $5.83\text{g}$.

Explanation of Solution

Reaction in the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{10}}$ is given as follows:

    ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{(s)}+2{\text{O}}_{\text{2}}\text{(g)}\to {\text{P}}_{\text{4}}{\text{O}}_{\text{10}}\text{(s)}$

The total moles of oxygen taken for the reaction is $\text{0}\text{.180}\text{mol}$.  In the formation of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$, the moles of oxygen that is reacted is $\text{0}\text{.140}\text{mol}$.  Therefore, the amount of oxygen that remains is calculated as follows:

    $\begin{array}{c}\text{Amount}\text{of}{\text{O}}_{\text{2}}\text{remaining}=0.180\text{mol}-0.140\text{mol}\\ =0.040\text{mol}\end{array}$

The amount of oxygen that is required to react with $\text{0}\text{.0465}\text{mol}$ of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is calculated as follows:

    $\begin{array}{c}{n}_{{\text{O}}_{\text{2}}}=\text{0}\text{.0465}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\times \frac{\text{2}\text{mol}{\text{O}}_{\text{2}}}{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}\\ =0.0930\text{mol}\end{array}$

As the remaining amount of oxygen is lesser than the amount of oxygen that is required for the reaction to complete, the limiting reactant is found to be oxygen and the excess reagent is found to be ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$.

Moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ consumed is calculated as follows:

    $\begin{array}{c}{n}_{{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}=\text{0}\text{.040}\text{mol}{\text{O}}_{\text{2}}\times \frac{\text{1}\text{mol}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}}{\text{2}\text{mol}{\text{O}}_{\text{2}}}\\ =0.020\text{mol}\end{array}$

Moles of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ that remains is calculated as follows:

    $\begin{array}{c}\text{Moles}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{remaining}=0.0465\text{mol}-0.020\text{mol}\\ =0.0265\text{mol}\end{array}$

Mass of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ that remains is calculated as follows:

    $\begin{array}{c}\text{Mass}\text{of}{\text{P}}_{\text{4}}{\text{O}}_{\text{6}}\text{remaining}=0.0265\text{mol}\times \text{220}\text{g}\cdot {\text{mol}}^{\text{-1}}\\ =5.83\text{g}\end{array}$

Therefore, the mass of excess reactant that is unreacted is $5.83\text{g}$.