Chapter 15, Problem 126AP

Interpretation Introduction

(a)

Interpretation:

The balanced molecular equation, complete ionic equation and net ionic equation for the given reaction mixture is to be written.

Concept Introduction:

A balanced chemical equation represents an equation in which all the reactants and products are written with their stoichiometric coefficient and physical states. The number of atoms of an element on the both sides of the equation is equal.

The complete ionic equation is the chemical equation of a reaction in which the ionic compounds are written in their dissociated ionic forms.

The net ionic equation is the one in which the common ions present on both the sides of the complete ionic reaction get eliminated and only the chemical species actually participating in the reaction is represented.

Answer to Problem 126AP

The balanced chemical reaction between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$3{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{a}\text{q}\right)+2{\text{FeCl}}_{\text{3}}\left(\text{a}\text{q}\right)\to 6{\text{NH}}_{\text{4}}\text{Cl}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{a}\text{q}\right)$

The complete ionic equation between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$\left(\begin{array}{l}6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+3{\text{S}}^{2-}\left(\text{a}\text{q}\right)+\\ 2{\text{Fe}}^{3+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)\end{array}\right)\to 6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The net ionic equation between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as:

$3{\text{S}}^{2-}\left(\text{a}\text{q}\right)+2{\text{Fe}}^{3+}\left(\text{a}\text{q}\right)\to {\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$.

Explanation of Solution

The unbalanced chemical reaction between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{a}\text{q}\right)+{\text{FeCl}}_{\text{3}}\left(\text{a}\text{q}\right)\to {\text{NH}}_{\text{4}}\text{Cl}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The number of atoms of elements on both sides of the equation is not equal. Therefore, multiply ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ by three, ${\text{FeCl}}_{\text{3}}$ by two, ${\text{NH}}_{\text{4}}\text{Cl}$ by six for the balanced chemical reaction.

The balanced chemical reaction between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$3{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{a}\text{q}\right)+2{\text{FeCl}}_{\text{3}}\left(\text{a}\text{q}\right)\to 6{\text{NH}}_{\text{4}}\text{Cl}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The complete ionic equation between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as:

$\left(\begin{array}{l}6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+3{\text{S}}^{2-}\left(\text{a}\text{q}\right)+\\ 2{\text{Fe}}^{3+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)\end{array}\right)\to 6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The ${\text{NH}}_4^{+}$ and ${\text{S}}^{2-}$ ion is common on the both sides of the complete ionic reaction therefore, to write net ionic reaction these ion must be eliminated from the reaction. The net ionic equation between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$3{\text{S}}^{2-}\left(\text{a}\text{q}\right)+2{\text{Fe}}^{3+}\left(\text{a}\text{q}\right)\to {\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$.

Interpretation Introduction

(b)

Interpretation:

The mass of precipitate formed on mixing given solutions of ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of moles of solute dissolved in one liter of the solution. The formula for molarity is given as,

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

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

The limiting reagent of a reaction is that reactant of the reaction that controls the amount of product formed. The limiting agents limits the amount of product and by adding some more amount of the limiting reagent in the reaction mixture, the amount of product can be increased.

Answer to Problem 126AP

The mass of precipitate formed on mixing the given solutions of ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is $1.73\text{g}$.

Explanation of Solution

The volume of $0.500\text{M}$ ammonium sulfide is $50.0\text{mL}$.

The volume of $\text{0}\text{.250}\text{M}\text{iron }\left(\text{III}\right)\text{ chloride}$ is $100.0\text{mL}$.

The molarity of a solution is given as:

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

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

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

$\text{n}=\text{M}\text{V}$    (1)

Substitute the value of molarity and volume of ammonium sulfide solution in equation (1).

$\begin{array}{c}\text{n}=\left(0.500\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(50.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.025\text{mol}\end{array}$

The number of moles of ammonium sulfide present in the reaction mixture is $0.025\text{mol}$.

Substitute the value of molarity and volume of $\text{iron }\left(\text{III}\right)\text{ chloride}$ solution in equation (1).

$\begin{array}{c}\text{n}=\left(0.250\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(100.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =0.025\text{mol}\end{array}$

The number of moles of $\text{iron }\left(\text{III}\right)\text{ chloride}$ present in the reaction mixture is $0.025\text{mol}$.

The balanced chemical reaction between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$3{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{a}\text{q}\right)+2{\text{FeCl}}_{\text{3}}\left(\text{a}\text{q}\right)\to 6{\text{NH}}_{\text{4}}\text{Cl}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

Three moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ reacts with two moles of ${\text{FeCl}}_{\text{3}}$ to produce six moles of ${\text{NH}}_{\text{4}}\text{Cl}$ and one mole of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$. Therefore, the relation between the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ and ${\text{FeCl}}_{\text{3}}$ reacts is given as:

${\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}=\frac{3{\text{n}}_{{\text{FeCl}}_{\text{3}}}}{2}$

Where,

• ${\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}$ represents the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$.
• ${\text{n}}_{{\text{FeCl}}_{\text{3}}}$ represents the number of moles of ${\text{FeCl}}_{\text{3}}$.

Substitute the value of number of moles of ${\text{FeCl}}_{\text{3}}$ in the above equation.

$\begin{array}{c}{\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}=\frac{\left(3\right)\left(0.025\text{mol}\right)}{2}\\ =0.0375\text{mol}\end{array}$

The number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ required to react with $0.025\text{mol}$ of ${\text{FeCl}}_{\text{3}}$ is $0.0375\text{mol}$ but the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ present in the reaction mixture is $0.025\text{mol}$, therefore ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is the limiting agent of the reaction. The amount of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ available will decide the amount of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$ produced.

The relation between the number of moles of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$ reacted and the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ produced is given as,

${\text{n}}_{{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}}=\frac{{\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}}{3}$

Where,

• ${\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}$ represents the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$.
• ${\text{n}}_{{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}}$ represents the number of moles of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$.

Substitute the value of the available number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ in the above equation.

$\begin{array}{c}{\text{n}}_{{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}}=\frac{\left(0.025\text{mol}\right)}{3}\\ =8.33\times {10}^{-3}\text{mol}\end{array}$

The number of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$ produced on mixing given solutions of ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is $8.33\times {10}^{-3}\text{mol}$.

The molar mass of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$ is $207.90\text{g}\cdot {\text{mol}}^{-1}$.

The relation between the number of moles and mass of a substance is given as,

$\text{m}=\text{n}{\text{M}}_{\text{m}}$

Where,

• $\text{n}$ represents the number of the substance.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the substance.

Substitute the value of $\text{n}$ and ${\text{M}}_{\text{m}}$ in the above equation for the mass of ${\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}$ produced.

$\begin{array}{c}\text{m}=\left(8.33\times {10}^{-3}\text{mol}\right)\left(207.90\text{g}\cdot {\text{mol}}^{-1}\right)\\ =1.73\text{g}\end{array}$

Therefore, the mass of precipitate formed on mixing given solutions of ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is $1.73\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The concentration of ammonium ion and $\text{iron}\left(\text{III}\right)$ ion left in solution after the completion of the reaction 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 as,

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

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 126AP

The concentration of ammonium ion and $\text{iron}\left(\text{III}\right)$ ion left in solution after the completion of the reaction is $\text{0}\text{.0084}\text{mol}$ and $0.056\text{M}$ respectively.

Explanation of Solution

The volume of $0.500\text{M}$ ammonium sulfide is $50.0\text{mL}$.

The volume of $\text{0}\text{.250}\text{M}\text{iron }\left(\text{III}\right)\text{ chloride}$ is $100.0\text{mL}$.

The number of moles of ammonium sulfide present in the reaction mixture is $0.025\text{mol}$.

The dissociation reaction of ammonium sulfide in aqueous solution is represented as,

${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}\left(\text{a}\text{q}\right)\to 2{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+{\text{S}}^{2-}\left(\text{a}\text{q}\right)$

Therefore, the relation between the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ and ${\text{NH}}_4^{+}$ is given as,

${\text{n}}_1=2{\text{n}}_2$

Where,

• ${\text{n}}_1$ represents the number of moles of ${\text{NH}}_4^{+}$.
• ${\text{n}}_2$ represents the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$.

Substitute the value of ${\text{n}}_2$ in above equation.

$\begin{array}{c}{\text{n}}_1=\left(2\right)\left(0.025\text{mol}\right)\\ =0.05\text{mol}\end{array}$

The complete ionic equation between the ammonium sulfide and $\text{iron }\left(\text{III}\right)\text{ chloride}$ is represented as,

$\left(\begin{array}{l}6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+3{\text{S}}^{2-}\left(\text{a}\text{q}\right)+\\ 2{\text{Fe}}^{3+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)\end{array}\right)\to 6{\text{NH}}_4^{+}\left(\text{a}\text{q}\right)+6{\text{Cl}}^{-}\left(\text{a}\text{q}\right)+{\text{Fe}}_{\text{2}}{\text{S}}_{\text{3}}\left(\text{s}\right)$

The number of moles of ammonium ions present in the solution before the reaction started and after the reaction completed is equal. Therefore, the number of moles of ammonium ion in the reaction after the completion of the reaction is $0.05\text{mol}$.

The total volume of the reaction mixture is given as,

$\text{V}={\text{V}}_1+{\text{V}}_2$

Where,

• ${\text{V}}_1$ represents the volume of ammonium sulfide solution.
• ${\text{V}}_2$ represents the volume of $\text{iron }\left(\text{III}\right)\text{ chloride}$ solution.

Substitute the value of ${\text{V}}_1$ and ${\text{V}}_2$ in the above equation.

$\begin{array}{c}\text{V}=50.0\text{mL}+100.0\text{mL}\\ =150.0\text{mL}\end{array}$

The molarity of a solution is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$    (2)

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Substitute the volume of the reaction mixture and number of moles of ammonium ion in equation (2).

$\begin{array}{c}\text{M}=\left(\frac{0.05\text{mol}}{150.0\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ =0.3333\text{M}\end{array}$

The molarity of ammonium ion after the reaction completed is $0.3333\text{M}$.

The relation between the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ and ${\text{FeCl}}_{\text{3}}$ is given as,

${\text{n}}_{{\text{FeCl}}_{\text{3}}}=\frac{2{\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}}{3}$

Where,

• ${\text{n}}_{{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}}$ represents the number of moles of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$.
• ${\text{n}}_{{\text{FeCl}}_{\text{3}}}$ represents the number of moles of ${\text{FeCl}}_{\text{3}}$.

Substitute the value of number of moles ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ of reacted in the above equation.

$\begin{array}{c}{\text{n}}_{{\text{FeCl}}_{\text{3}}}=\frac{\left(2\right)\left(0.025\text{mol}\right)}{3}\\ =0.0166\text{mol}\end{array}$

The number of moles of ${\text{FeCl}}_{\text{3}}$ required to react with $0.025\text{mol}$ of ${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is $0.0166\text{mol}$ but the number of moles of ${\text{FeCl}}_{\text{3}}$ present in the reaction mixture is $0.025\text{mol}$ therefore some amount of ${\text{FeCl}}_{\text{3}}$ will remain unreacted in the reaction mixture. The amount of ${\text{FeCl}}_{\text{3}}$ remain unreacted is given as,

$\text{n}={\text{n}}_1-{\text{n}}_2$

Where,

• ${\text{n}}_1$ represents the number of moles of ${\text{FeCl}}_{\text{3}}$ present before the reaction gets completed.
• ${\text{n}}_2$ represents the number of moles of ${\text{FeCl}}_{\text{3}}$ consumed in the reaction.

$$

Substitute the value of ${\text{n}}_1$ and ${\text{n}}_2$ in the above equation.

$\begin{array}{c}\text{n}=0.025\text{mol}-0.0166\text{mol}\\ =\text{0}\text{.0084}\text{mol}\end{array}$

The number of moles of ${\text{FeCl}}_{\text{3}}$ remains in the solution is $\text{0}\text{.0084}\text{mol}$. Therefore, the number of moles of ${\text{Fe}}^{\text{3+}}$ is $\text{0}\text{.0084}\text{mol}$.

Substitute the volume of the reaction mixture and number of moles of $\text{iron}\left(\text{III}\right)$ ion in equation (2).

$\begin{array}{c}\text{M}=\left(\frac{\text{0}\text{.0084}\text{mol}}{150.0\text{mL}}\right)\left(\frac{1000\text{mL}}{1\text{L}}\right)\left(\frac{1\text{M}}{1\text{mol}\cdot {\text{L}}^{-1}}\right)\\ =0.056\text{M}\end{array}$

The molarity of $\text{iron}\left(\text{III}\right)$ ion after the reaction completed is $0.056\text{M}$.

Interpretation Introduction

(d)

Interpretation:

The mass of solid $\text{iron }\left(\text{III}\right)\text{ chloride}$ required to prepare the given solution of $\text{iron }\left(\text{III}\right)\text{ chloride}$ 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 as,

$\text{M}=\frac{\text{m}}{{\text{M}}_{\text{m}}\text{V}}$

Where,

• $\text{m}$ represents the mass of the solute.
• $\text{V}$ represents the volume of the solution.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the solute.

Answer to Problem 126AP

The mass of $\text{iron }\left(\text{III}\right)\text{ chloride}$ required to prepare $100.0\text{mL}$ of $\text{0}\text{.250}\text{M}$ solution is $5.1975\text{g}$.

Explanation of Solution

The volume of $\text{0}\text{.250}\text{M}\text{iron }\left(\text{III}\right)\text{ chloride}$ is $100.0\text{mL}$.

The molar mass of $\text{iron }\left(\text{III}\right)\text{ chloride}$ is $207.90\text{g}\cdot {\text{mol}}^{-1}$.

The molarity of a solution is given as,

$\text{M}=\frac{\text{m}}{{\text{M}}_{\text{m}}\text{V}}$

Where,

• $\text{m}$ represents the mass of the solute.
• $\text{V}$ represents the volume of the solution.
• ${\text{M}}_{\text{m}}$ represents the molar mass of the solute.

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

$\text{m}=\text{M}{\text{M}}_{\text{m}}\text{V}$

Substitute the value of $\text{M}$, $\text{V}$ and ${\text{M}}_{\text{m}}$ in the above equation.

$\begin{array}{c}\text{m}=\left(\text{0}\text{.250}\text{M}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\left(207.90\text{g}\cdot {\text{mol}}^{-1}\right)\left(100.0\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\\ =5.1975\text{g}\end{array}$

Therefore, the mass of $\text{iron }\left(\text{III}\right)\text{ chloride}$ required to prepare $100.0\text{mL}$ of $\text{0}\text{.250}\text{M}$ solution is $5.1975\text{g}$.