Chapter F, Problem M.16E

(a)

Interpretation Introduction

Interpretation:

Balanced equation for the reaction given has to be written.

    ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Explanation of Solution

Skeletal equation is given as follows.

    ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing copper atoms:

In the above equation, there is one copper atom on the left side while three copper atoms are present in the right side of equation.  Adding coefficient $3$ before ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ in the reactant side balances the copper atoms on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing sulfate ions:

In the above equation, there are three sulfate ions on the left side of equation while there is one sulfate ion on the product side of equation.  Adding coefficient $3$ before ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ balances the sulfate ions on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing phosphorus atoms:

In the above equation, there is one phosphorus atom on the left side of equation while there are two phosphorus atoms on the product side of equation.  Adding coefficient $2$ before ${\text{PH}}_{\text{3}}$ balances the phosphorus atoms on both sides of the equation.  The chemical equation obtained is given as shown below;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+2{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+{\text{H}}_{\text{2}}\text{O}(l)$

Balancing water molecules:

In the above equation, there are fifteen water molecules on the left side of equation while there is one water molecule on the product side of equation.  Adding coefficient $15$ before ${\text{H}}_{\text{2}}\text{O}$ balances the water molecules on both sides of the equation.  The balanced chemical equation obtained is given as shown below;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+2{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+15{\text{H}}_{\text{2}}\text{O}(l)$

(b)

Interpretation Introduction

Interpretation:

Name of the reactants and products in the reaction has to be given.

Explanation of Solution

The balanced chemical equation is given as follows;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+2{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+15{\text{H}}_{\text{2}}\text{O}(l)$

Name of the reactants:

The reactant ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is named as copper(II) sulfate pentahydrate and ${\text{PH}}_{\text{3}}$ is named as phosphine.

Names of the products:

The product ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ is named as copper triphosphide, ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ is named as sulfuric acid and ${\text{H}}_{\text{2}}\text{O}$ is named as water.

(c)

Interpretation Introduction

Interpretation:

Limiting reactant in the reaction has to be determined.

Explanation of Solution

The balanced chemical equation is given as follows;

    ${\text{3CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}(\text{s})+2{\text{PH}}_{\text{3}}(\text{g})\to {\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}(\text{s})+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}(aq)+15{\text{H}}_{\text{2}}\text{O}(l)$

Mass of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ taken is $\text{0}\text{.110}\text{kg}$ and that of phosphine is $\text{4}\text{.94}\text{g}$ that is $85.0\%$.  The molar mass of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $249.68\text{g}\cdot {\text{mol}}^{-1}$.  The number of moles of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ and ${\text{PH}}_{\text{3}}$ is calculated as follows;

    $\begin{array}{c}{n}_{({\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O})}=\frac{110\text{g}}{249.68\text{g}\cdot {\text{mol}}^{-1}}\\ =0.44\text{mol}\\ n_{(P{H}_3)}=\frac{4.94\text{g}}{33.99\text{g}\cdot {\text{mol}}^{-1}}\times \frac{85}{100}\\ =0.123\text{mol}\end{array}$

From the balanced chemical equation, it is found that three moles of ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ reacts with two moles of phosphine to produce one mole of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$.  The amount of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ can be calculated as shown below;

    $\begin{array}{c}n({\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}})=0.44\text{mol of}{\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}\times \frac{1\text{mol}\text{of}{\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}}{\text{3}\text{mol}\text{of}{\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}}\\ =0.147\text{mol of}{\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}\\ n({\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}})=0.123\text{mol of}{\text{PH}}_{\text{3}}\times \frac{1\text{mol}\text{of}{\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}}{\text{2}\text{mol}\text{of}{\text{PH}}_{\text{3}}}\\ =0.062\text{mol of}{\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}\end{array}$

From the above calculation it is found that phosphine is the limiting reactant while ${\text{CuSO}}_{\text{4}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is the excess reagent.

(d)

Interpretation Introduction

Interpretation:

Mass of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ produced if the percentage yield is $6.31\%$ has to be calculated.

Explanation of Solution

The limiting reactant is found to the ${\text{PH}}_{\text{3}}$ and the moles of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ that is formed is calculated to be $\text{0}\text{.062}\text{mol}$.  Thus the theoretical yield of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ formed can be calculated as follows.

    $\begin{array}{c}\text{Theoretical yield}=\text{0}\text{.062}\text{mol}\times 252.56\text{g}\cdot {\text{mol}}^{-1}\\ =15.66\text{g}\end{array}$

The percentage yield of reaction is given as $6.31\%$.

Actual yield can be calculated as follows.

  $\begin{array}{c}\text{Actual yield}=\frac{\text{Percentage yield}\times \text{Theoretical yield}}{100}\\ =\frac{6.31\times 15.66\text{g}}{100}\\ =0.99\text{g}\end{array}$

Thus the mass of ${\text{Cu}}_{\text{3}}{\text{P}}_{\text{2}}$ produced is $0.99\text{g}$.