Chapter 4, Problem 4.132AP

(a)

Interpretation Introduction

Interpretation: The questions corresponding to acidic mine drainage are to be answered.

Concept introduction: The number of given moles of solute divided by a liter of solution is termed as molarity. It can be given by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\left(n\right)}{\text{Volume}\text{of}\text{solution}\left(V\right)\text{in}\text{L}}$

To determine: The molarity of iron and zinc in the drainage.

Answer to Problem 4.132AP

Solution

The molarity of iron is $\underset{\_}{1.43M}$ and the molarity of zinc in the drainage is $\underset{\_}{9.17\times 10^{-2}M}$.

Explanation of Solution

Explanation

The mass $\left(g\right)$ of iron dissolved in one liter of drainage is $80.0\text{g}$.

The molar mass of iron $\left(\text{Fe}\right)$ is $55.85\text{g/mol}$.

The mass $\left(g\right)$ of zinc dissolved in one liter of drainage is $6.0\text{g}$.

The molar mass of iron $\left(\text{Fe}\right)$ is $65.41\text{g/mol}$.

The volume of the drainage is $1\text{L}$.

Firstly, the moles of iron and zinc are calculated that will be used in the formula of molarity further.

The moles of iron $\left(\text{Fe}\right)$ and zinc $\left(\text{Zn}\right)$ are calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

Substitute the values of mass and molar mass of iron $\left(\text{Fe}\right)$ in the above formula as,

$\begin{array}{c}\text{Moles}\text{of}\text{iron}\left(\text{Fe}\right)=\frac{80.0\text{g}}{55.85\text{g}}\times 1\text{mol}\text{Fe}\\ =1.43\text{mol}\text{Fe}\end{array}$

Substitute the values of mass and molar mass of zinc $\left(\text{Zn}\right)$ in the above formula as,

$\begin{array}{c}\text{Moles}\text{of}\text{zinc}\left(\text{Zn}\right)=\frac{6.0\text{g}}{65.41\text{g}}\times 1\text{mol}\text{Zn}\\ =9.17\times {10}^{-2}\text{mol}\text{Zn}\end{array}$

The molarity of iron and zinc in the drainage is calculated by the formula,

$\text{Molarity}=\frac{\text{Number}\text{of}\text{moles}\text{of}\text{solute}\left(n\right)}{\text{Volume}\text{of}\text{solution}\left(V\right)\text{in}\text{L}}$

Substitute the values of moles and volume for iron in the above formula of molarity as,

$\begin{array}{c}\text{Molarity}\text{of}\text{Fe}=\frac{1.43\text{mol}\text{Fe}}{1\text{L}}\\ =1.43\text{mol/L}\\ \text{=}\underset{\_}{1.43M}\end{array}$

Substitute the values of moles and volume for zinc in the above formula of molarity as,

$\begin{array}{c}\text{Molarity}\text{of}\text{Zn}=\frac{9\times {10}^{-2}\text{mol}\text{Zn}}{1\text{L}}\\ =9.17\times {10}^{-2}\text{mol/L}\\ \text{=}\underset{\_}{9.17\times 10^{-2}M}\end{array}$

Therefore, the molarity of iron is $\underset{\_}{1.43M}$ and the molarity of zinc in the drainage is $\underset{\_}{9.17\times 10^{-2}M}$.

(b)

Interpretation Introduction

To determine: The complete chemical equation and a net ionic equation for the given process.

Answer to Problem 4.132AP

Solution

The complete and balanced equation for the process is,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The net ionic equation for the given process is,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

Explanation

The given in complete equation is,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to$

The unbalanced equation is written as,

$\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

To balance the element iron in the above equation a coefficient of $2$ is placed in front of $\text{Fe}$ to the reactant side of the equation as,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The oxygen and hydrogen atoms are balanced by placing coefficient of $3$ in front of ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ to the reactant side and a coefficient of $6$ in front of water molecule to the product side of the equation as,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The complete and balanced equation is,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The net ionic equation of the above complete and balanced equation is obtained as follows.

Firstly, a balanced molecular equation is to be written that gives the overall ionic equation as,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\left(aq\right){\text{+3SO}}_{\text{4}}{}^{2-}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+{\text{3SO}}_{\text{4}}{}^{2-}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The elimination of the spectator ions such as, ${\text{3SO}}_{\text{4}}{}^{2-}\left(aq\right)$ from the overall ionic equation will produce the net ionic equation as,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Therefore, the net ionic equation is,

$\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

(c)

Interpretation Introduction

To determine: The balanced net ionic equation for the reaction between smithsonite and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that produces ${\text{Zn}}^{\text{2+}}\left(aq\right)$.

Answer to Problem 4.132AP

Solution

The balanced net ionic equation for the reaction between smithsonite and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that produces ${\text{Zn}}^{\text{2+}}\left(aq\right)$ is,

${\text{ZnCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Explanation of Solution

Explanation

The reaction between smithsonite $\left({\text{ZnCO}}_{\text{3}}\right)$  and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ produces zinc sulfate $\left({\text{ZnSO}}_4\left(aq\right)\right)$, carbon dioxide and water as,

${\text{ZnCO}}_{\text{3}}\left(s\right)+{\text{H}}_2{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{ZnSO}}_4\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The above equation is balanced. So there is no need to balance the atoms in the equation.

The net ionic equation of the above complete and balanced equation is obtained as follows.

Firstly, a balanced molecular equation is to be written that gives the overall ionic equation as,

${\text{ZnCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)+{\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{SO}}_4{}^{2-}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

The elimination of the spectator ions such as, ${\text{SO}}_{\text{4}}{}^{2-}\left(aq\right)$ from the overall ionic equation will produce the net ionic equation as,

${\text{ZnCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

Therefore, the net ionic equation is,

${\text{ZnCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

(d)

Interpretation Introduction

To determine: The value of $x$ in the given formula ${\text{Zn}}_x{\text{Fe}}_{1-x}$ of the mineral in the deposit.

Answer to Problem 4.132AP

Solution

The value of $x$ in the given formula ${\text{Zn}}_x{\text{Fe}}_{1-x}$ ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ of the mineral is $\underset{\_}{0.18}$.

Explanation of Solution

Explanation

The given formula is,

${\text{Zn}}_x{\text{Fe}}_{1-x}$ ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$.

By manipulating the above formula becomes as,

${\text{Zn}}_x{\text{Fe}}_{3-x}{\text{O}}_{\text{4}}$

The mole-mole conversion of the above formula is obtained by dividing moles of iron by mole of zinc as,

$\frac{\left(3-x\right)\text{mol}\text{Fe}}{x\text{mol}\text{Zn}}$ (1)

The moles of iron $\left(\text{Fe}\right)$ and zinc $\left(\text{Zn}\right)$ are calculated by the formula,

$\text{Moles}=\frac{\text{Mass}\left(\text{g}\right)}{\text{Molar}\text{mass}}$

Substitute the values of mass and molar mass of iron $\left(\text{Fe}\right)$ in the above formula as,

$\begin{array}{c}\text{Moles}\text{of}\text{iron}\left(\text{Fe}\right)=\frac{80.0\text{g}}{55.85\text{g}}\times 1\text{mol}\text{Fe}\\ =1.43\text{mol}\text{Fe}\end{array}$

Substitute the values of mass and molar mass of zinc $\left(\text{Zn}\right)$ in the above formula as,

$\begin{array}{c}\text{Moles}\text{of}\text{zinc}\left(\text{Zn}\right)=\frac{6.0\text{g}}{65.41\text{g}}\times 1\text{mol}\text{Zn}\\ =9.17\times {10}^{-2}\text{mol}\text{Zn}\end{array}$

Substitute the values of moles of iron and zinc in equation (1) as,

$\begin{array}{c}9.17\times {10}^{-2}\text{mol}\text{Zn}\times \frac{\left(3-x\right)\text{mol}\text{Fe}}{\text{x}\text{mol}\text{Zn}}=1.43\text{mol}\text{Fe}\\ \text{=}\left(9.17\times {10}^{-2}\right)\left(3-x\right)=1.43x\end{array}$

The value of $x$ is calculated as,

$\begin{array}{c}3\left(9.17\times {10}^{-2}\right)-x\left(9.17\times {10}^{-2}\right)=1.43x\\ \left(1.43+9.17\times {10}^{-2}\right)x=3\left(9.17\times {10}^{-2}\right)\\ x=\frac{0.2751}{1.5217}\\ x=\underset{\_}{0.18}\end{array}$

Therefore, the value of $x$ in the given formula ${\text{Zn}}_x{\text{Fe}}_{1-x}$ ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ of the mineral is $\underset{\_}{0.18}$.

Conclusion

1. a. The molarity of iron is $\underset{\_}{1.43M}$ and the molarity of zinc in the drainage is $\underset{\_}{9\times 10^{-2}M}$.
2. b. The complete and balanced equation for the process is,

   $\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+3{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\left(aq\right)\to {\text{Fe}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

   The net ionic equation for the given process is,

   $\text{2Fe}{\left(\text{OH}\right)}_{\text{3}}\left(s\right)+6{\text{H}}^{+}\left(aq\right)\to 2{\text{Fe}}^{3+}\left(aq\right)+6{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

3. c. The balanced net ionic equation for the reaction between smithsonite and ${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$ that produces ${\text{Zn}}^{\text{2+}}\left(aq\right)$ is,

   ${\text{ZnCO}}_{\text{3}}\left(s\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{Zn}}^{2+}\left(aq\right)+{\text{CO}}_{\text{2}}\left(g\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)$.

4. d. The value of $x$ in the given formula ${\text{Zn}}_x{\text{Fe}}_{1-x}$ ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ of the mineral is $\underset{\_}{0.18}$