Chapter 4, Problem 4.96P

(a)

Interpretation Introduction

Interpretation: The redox reaction needs to be identified.

Concept Introduction: A reaction is said to be a redox reaction if both oxidation and reduction takes place. In the reduction oxidation state of the atom decreases and in oxidation, the oxidation state of the atom increases.

Explanation of Solution

The given two reactions are as follows:

(1) $6\text{HCl}\left(aq\right)+{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\to 2{\text{FeCl}}_{\text{3}}\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

(2) $\text{2HCl}\left(aq\right)+\text{Fe}\left(s\right)\to {\text{FeCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

In reaction (1), on the reactant side, the oxidation state of H and Cl in HCl is +1 and -1 respectively. The oxidation state of Fe and O in ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ is +3 and -2 respectively.

Now, in the product side, the oxidation state of Fe and Cl in ${\text{FeCl}}_{\text{3}}$ is +3 and -3 respectively and that of H and O in ${\text{H}}_{\text{2}}\text{O}$ is +1 and -2 respectively.

Since, there is no change in any oxidation state of element thus, it is not a redox reaction.

In reaction (2), in the reactant side, the oxidation state of H and Cl in HCl is +1 and -1 respectively. The oxidation state of Fe is 0.

In the product side, the oxidation state of Fe and Cl in ${\text{FeCl}}_2$ is +2 and -1 respectively and that of H in H₂ is 0.

Now, oxidation number of Fe increases from 0 to +2 and that of H is decreased from +1 to 0 thus, reaction (2) is a redox reaction.

(b)

Interpretation Introduction

Interpretation: The mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ removed and ${\text{FeCl}}_{\text{3}}$ produced needs to be determined.

Concept Introduction: The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

The equation for the washing steel in hydrochloric acid is represented as follows:

  $6\text{HCl}\left(aq\right)+{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\to 2{\text{FeCl}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The concentration of HCl is 3 M and volume is $2.50\times {10}^3\text{ L}$ thus, number of moles of HCl can be calculated as follows:

  $n_{\text{HCl}}=2.50\times {10}^3\text{ L}\times \frac{3\text{ mol HCl}}{1\text{ L}}=7.50\times {10}^3\text{ mol HCl}$ n

From the above equation, 6 mol of HCl required 1 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ thus, the number of moles of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ using mass of HCl can be calculated as follows:

  $n_{{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}}=7.50\times {10}^3\text{ mol HCl}\times \frac{1\text{ mol Fe}\text{O}{}_{\text{3}}}{6\text{ mol HCl}}=1.25\times {10}^3{\text{ mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}$

The molar mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is 159.70 g/mol thus, mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ can be calculated as follows:

  $m=n\times M=\left(1.25\times {10}^3\text{ mol}\right)\left(159.70\text{ g/mol}\right)=2\times {10}^5\text{ g}$

Thus, mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}$ is $2\times {10}^5\text{ g}$ .

Also, 6 moles of HCl produces 2 mol of ${\text{FeCl}}_{\text{3}}$ thus, number of moles of ${\text{FeCl}}_{\text{3}}$ using number of moles of HCl can be calculated as follows:

  $n_{{\text{FeCl}}_{\text{3}}}=7.5\times {10}^3\left(\frac{2}{6}\right)\text{ mol=2}\text{.50}\times {\text{10}}^3\text{ mol}$

Now, molar mass of ${\text{FeCl}}_{\text{3}}$ is 162.20 g/mol thus, mass of ${\text{FeCl}}_{\text{3}}$ can be calculated as follows:

  $m=n\times M=\left(2.50\times {10}^3\text{ mol}\right)\left(162.20\text{ g/mol}\right)=4.06\times {10}^5\text{ g}$

(c)

Interpretation Introduction

Interpretation: The mass of Fe lost and FeCl₂ produced needs to be determined.

Concept Introduction: The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

The reaction (2) is as follows:

  $\text{2HCl}\left(aq\right)+\text{Fe}\left(s\right)\to {\text{FeCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

2 mol of HCl reacts with 1 mol of Fe thus, number of moles of Fe reacting with given number of moles of HCl will be:

  $n=7.50\times {10}^3\text{ mol}\left(\frac{1}{2}\right)=3.75\times {10}^3\text{ mol}$

Molar mass of Fe is 55.845 g/mol thus, mass of Fe will be:

  $m=n\times M=3.75\times {10}^3\text{ mol}\times 55.845\text{ g/mol=2}\text{.09}\times {\text{10}}^5\text{ g}$

Here, 2 mol of HCl gives 1 mol of ${\text{FeCl}}_2$ thus, the number of moles of ${\text{FeCl}}_2$ from number of moles of HCl can be calculated as follows:

  $n=7.50\times {10}^3\text{ mol}\left(\frac{1}{2}\right)=3.75\times {10}^3\text{ mol}$

Molar mass of ${\text{FeCl}}_2$ is 126.75 g/mol thus, mass of ${\text{FeCl}}_2$ will be:

  $m=n\times M=\left(3.75\times {10}^3\text{ mol}\right)\left(126.75\text{ g/mol}\right)=4.75\times {10}^5\text{ g}$

(d)

Interpretation Introduction

Interpretation: The mass ratio of FeCl₂ and FeCl₃ needs to be determined.

Concept Introduction: The number of moles can be calculated as follows:

  $n=\frac{m}{M}$

Here, m is mass and M is molar mass.

Explanation of Solution

The equation for the washing steel in hydrochloric acid is represented as follows:

  $6\text{HCl}\left(aq\right)+{\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)\to 2{\text{FeCl}}_3\left(aq\right)+3{\text{H}}_{\text{2}}\text{O}\left(l\right)$

Let the mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ be 1 g thus, from the equation 1 mol of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ produces 2 mol of ${\text{FeCl}}_3\left(aq\right)$ thus, the number of moles of ${\text{FeCl}}_3\left(aq\right)$ using the mass of ${\text{Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(s\right)$ can be calculated as follows:

  $n=1{\text{ g Fe}}_{\text{2}}{\text{O}}_{\text{3}}\left(\frac{1{\text{mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}{159.70{\text{ g Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\right)\left(\frac{{\text{2mol FeCl}}_{\text{3}}}{{\text{1mol Fe}}_{\text{2}}{\text{O}}_{\text{3}}}\right)=1.25\times {10}^{-2}{\text{ mol FeCl}}_{\text{3}}$

The molar mass of ${\text{ FeCl}}_{\text{3}}$ is 162.20 g/mol thus, mass of ${\text{ FeCl}}_{\text{3}}$ can be calculated as follows:

  $m=1.25\times {10}^{-2}\text{ mol }\times 162.20\text{ g/mol=2}\text{.03}g$

The second reaction is as follows:

  $\text{2HCl}\left(aq\right)+\text{Fe}\left(s\right)\to {\text{FeCl}}_2\left(aq\right)+{\text{H}}_{\text{2}}\left(g\right)$

The mass of Fe is 0.280 g. thus, number of moles will be:

  $n=\frac{m}{M}=\frac{0.280\text{ g}}{55.85\text{ g/mol}}=0.0050\text{ mol}$

From chemical reaction, 1 mol of Fe gives 1 mol of ${\text{FeCl}}_2\left(aq\right)$ thus, number of moles of ${\text{FeCl}}_2\left(aq\right)$ produced from 0.0050 mol of Fe will be 0.0050 mol.

Molar mass of ${\text{FeCl}}_2\left(aq\right)$ is 126.75 g/mol thus, mass of ${\text{FeCl}}_2\left(aq\right)$ will be:

  $m=n\times M=\left(0.0050\text{ mol}\right)\left(126.75\text{ g/mol}\right)=0.634\text{ g}$

Thus, the mass ratio of ${\text{FeCl}}_2\left(aq\right)$ to ${\text{FeCl}}_3\left(aq\right)$ will be:

  $\frac{m_{{\text{FeCl}}_2\left(aq\right)}}{m_{{\text{FeCl}}_3\left(aq\right)}}=\frac{0.635\text{ g}}{2.03\text{ g}}=0.313$

Thus, the mass ratio is 0.313.