Chapter 12, Problem 12.27P

Interpretation Introduction

Interpretation:

Mass of metallic copper formed along with the concentration of various ions after the completion of reaction has to be calculated.

Concept Introduction:

The formula to evaluate moles from molarity is given as follows:

  $n=MV$

Here,

$M$ denotes molarity.

$n$ denotes number of moles.

$V$ denotes volume.

Answer to Problem 12.27P

Mass of copper formed is $3.9319\text{ g}$. Concentration of ${\text{SO}}_4{}^{2-}$ and ${\text{Zn}}^{2+}$.is $\text{0}\text{.165 mol/L}$ and $1.0584\text{ mol/L}$ respectively.

Explanation of Solution

The conversion factor to convert liters to milliliters is as follows:

  $\text{1 L}=\text{1000 mL}$

Hence convert $\text{375 mL}$ to $\text{L}$ as follows:

  $\begin{array}{c}\text{Volume}=\left(\text{375 mL}\right)\left(\frac{1\text{ L}}{1000\text{ mL}}\right)\\ =\text{0}\text{.375 L}\end{array}$

The formula to convert mass in gram to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (2)

Substitute $\text{30 g}$ for mass and $\text{65}\text{.38 g/mol}$ for molar mass to calculate the number of moles of zinc in equation (2).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{30 g}}{\text{65}\text{.38 g/mol}}\\ =0.458785\text{ mol}\end{array}$

The formula to evaluate moles from molarity is given as follows:

  $n=MV$        (1)

Substitute $\text{0}\text{.165 mol/L}$ for $M$ and $0.375\text{ L}$ for $V$ in equation (1) to calculate moles of ${\text{CuSO}}_{\text{4}}$.

  $\begin{array}{c}n=\left(\text{0}\text{.165 mol/L}\right)\left(0.375\text{ L}\right)\\ =0.061875\text{ mol}\end{array}$

The balanced chemical reaction is given as follows:

  $\text{Zn}\left(s\right)+{\text{CuSO}}_{\text{4}}\left(aq\right)\to {\text{ZnSO}}_{\text{4}}\left(aq\right)+\text{Cu}\left(\text{s}\right)$        (2)

In accordance with stoichiometry of the balanced equation (2), one mole of ${\text{CuSO}}_{\text{4}}$ combines with one mole metallic zinc, hence, as ${\text{CuSO}}_{\text{4}}$ available is only $0.061875\text{ mol}$ it must be the limiting reagent. Thus moles of $\text{Cu}\left(\text{s}\right)$ formed is $0.061875\text{ mol}$.

The formula to convert mass in gram to moles is as follows:

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{Molar mass}\right)$        (3)

Substitute $0.061875\text{ mol}$ for moles and $\text{63}\text{.546 g/mol}$ for molar mass of $\text{Cu}\left(\text{s}\right)$ in equation (3).

  $\begin{array}{c}\text{Mass}=\left(0.061875\text{ mol}\right)\left(\text{63}\text{.546 g/mol}\right)\\ =3.9319\text{ g}\end{array}$

Since all the ${\text{Cu}}^{2+}$ present in ${\text{CuSO}}_{\text{4}}$ is consumed and is converted to precipitate $\text{Cu}$, Ionic species present in solution after reaction include ${\text{SO}}_4{}^{2-}$ and ${\text{Zn}}^{2+}$.

Moles of ${\text{SO}}_4{}^{2-}$ present in $\text{0}{\text{.165 mol/L CuSO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Moles of SO}}_4^{2-}=\left(0.375\text{ L}\right)\left(\frac{\text{0}{\text{.165 mol CuSO}}_{\text{4}}}{1\text{ L}}\right)\left(\frac{{\text{1 mol SO}}_4^{2-}}{{\text{1 mol CuSO}}_{\text{4}}}\right)\\ =0.061875\text{ mol}\end{array}$

Thus, concentration of ${\text{SO}}_4{}^{2-}$ is calculated as follows:

  $\begin{array}{c}{\text{Concentration of SO}}_4^{2-}=\frac{0.061875\text{ mol}}{0.375\text{ L}}\\ =\text{0}\text{.165 mol/L}\end{array}$

Thus, ${\text{Zn}}^{2+}$ present in excess is calculated as follows:

  $\begin{array}{c}\text{Excess moles of Zn}=\text{Total moles of Zn}-\text{Moles of Zn reacted}\\ =\left(0.458785\text{ mol Zn}\right)-\left(0.061875\text{ mol}\right)\\ =0.39691\text{ mol Zn}\end{array}$

Hence, concentration of ${\text{Zn}}^{2+}$ present after the reaction is calculated as follows:

  $\begin{array}{c}{\text{Concentration of Zn}}^{2+}=\frac{\left(0.39691\text{ mol}\right)}{0.375\text{ L}}\\ =1.0584\text{ mol/L}\end{array}$