Chapter 16, Problem 145E

Interpretation Introduction

Interpretation:

The mass of iron $\left(\text{III}\right)$ oxide that could be collected from $35.0\text{mL}$ of $0.516\text{M}$ iron $\left(\text{III}\right)$ nitrate is to be calculated.

Concept introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given by the expression shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

The relation between number of moles and mass of a substance is given by the expression shown below.

$\text{n}=\frac{\text{m}}{\text{Molar}\text{mass}}$

Answer to Problem 145E

The mass of iron $\left(\text{III}\right)$ oxide could be collected from $35.0\text{mL}$ of $0.516\text{M}$ iron $\left(\text{III}\right)$ nitrate is $1.44\text{g}$.

Explanation of Solution

The molarity of iron $\left(\text{III}\right)$ nitrate solution is $0.516\text{M}$.

The volume of the $0.516\text{M}$ iron $\left(\text{III}\right)$ nitrate solution is $35.0\text{mL}$.

The conversion of volume in $\text{L}$ is shown below.

$\begin{array}{rll}\text{V}& =& \left(35.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ & =& 35.0\times {10}^{-3}\text{L}\end{array}$

The molar mass of iron $\left(\text{III}\right)$ oxide $\left({\text{Fe}}_2{\text{O}}_{\text{3}}\right)$ is $159.70\text{g}/\text{mol}$.

The equation for the sequence reaction for the conversion of iron $\left(\text{III}\right)$ nitrate into iron $\left(\text{III}\right)$ hydroxide and then finally into iron $\left(\text{III}\right)$ oxide is shown below.

$2\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\to 2\text{Fe}{\left(\text{OH}\right)}_{\text{3}}\to {\text{Fe}}_2{\text{O}}_{\text{3}}$

The molarity of a solution is given by the expression as shown below.

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ is the number of moles of the solute.

• $\text{V}$ is the volume of the solution.

Rearrange the above equation for the value of $\text{n}$.

$\text{n}=\text{MV}$

Substitute the value of molarity and volume of iron $\left(\text{III}\right)$ nitrate solution in the above expression.

$\begin{array}{rll}\text{n}& =& \left(0.516\text{M}\right)\left(\frac{1\text{mol}/\text{L}}{1\text{M}}\right)\left(35.0\times {10}^{-3}\text{L}\right)\\ & =& 18.06\times {10}^{-3}\text{mol}\end{array}$

The number of moles of iron $\left(\text{III}\right)$ nitrate present in the solution is $18.06\times {10}^{-3}\text{mol}$.

Two moles of iron $\left(\text{III}\right)$ nitrate produce one mole of iron $\left(\text{III}\right)$ oxide. Therefore, the relation between the number of moles of iron $\left(\text{III}\right)$ nitrate and iron $\left(\text{III}\right)$ oxide is given by the expression as shown below.

${\text{n}}_{{\text{Fe}}_2{\text{O}}_{\text{3}}}=\frac{{\text{n}}_{\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}}{2}$…(1)

Where,

• ${\text{n}}_{\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}$ is the number of moles of $\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$.

• ${\text{n}}_{{\text{Fe}}_2{\text{O}}_{\text{3}}}$ is the number of moles of ${\text{Fe}}_2{\text{O}}_{\text{3}}$.

Substitute the value of ${\text{n}}_{\text{Fe}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}}$ in the equation (1).

$\begin{array}{rll}{\text{n}}_{{\text{Fe}}_2{\text{O}}_{\text{3}}}& =& \frac{18.06\times {10}^{-3}\text{mol}}{2}\\ & =& 9.03\times {10}^{-3}\text{mol}\end{array}$

The relation between number of moles and mass of a substance is given by the expression as shown below.

$\text{m}=\text{n}\times \text{Molar}\text{mass}$

Where,

• $\text{m}$ is the mass of the substance.

• $\text{n}$ is the number of moles of the substance.

Substitute the value of number of moles and molar mass of ${\text{Fe}}_2{\text{O}}_{\text{3}}$ in the above equation.

$\begin{array}{rll}\text{m}& =& \left(9.03\times {10}^{-3}\text{mol}\right)\left(159.70\text{g}/\text{mol}\right)\\ & =& 1.442\text{g}\\ & \approx & 1.44\text{g}\end{array}$

Therefore, the mass of iron $\left(\text{III}\right)$ oxide could be collected from $35.0\text{mL}$ of $0.516\text{M}$ iron $\left(\text{III}\right)$ nitrate is $1.44\text{g}$.

Conclusion

The mass of iron $\left(\text{III}\right)$ oxide could be collected from $35.0\text{mL}$ of $0.516\text{M}$ iron $\left(\text{III}\right)$ nitrate is $1.44\text{g}$.