Chapter 4, Problem 82GQ

The reaction of iron metal and chlorine gas to give iron(III) chloride is illustrated below.

(a) Write the balanced chemical equation for the reaction.

(b) Beginning with 10.0 g of iron, what mass of Cl₂, in grams, is required for complete reaction? What mass of FeCl₃ can be produced?

(c) If only 18.5 g of FeCl₃ is obtained from 10.0 g of iron and excess Cl₂, what is the percent yield?

(d) If 10.0 g each of iron and chlorine are combined, what is the theoretical yield of iron(III) chloride?

Interpretation Introduction

Interpretation:

Balanced chemical equation for the reaction of iron metal with chlorine gas forming iron(III) chloride has to be written.

Concept introduction:

• For chemical reaction balanced chemical reaction equation written in accordance with the Law of conservation of mass.
• Law of conservation of mass states that for a reaction total mass of the reactant and product must be equal.

Answer to Problem 82GQ

The balanced chemical reaction equation for this reaction is,

  $2F{e}_{(s)}+3C{l}_2{}_{(g)}\stackrel{}{\to }2FeC{l}_3{}_{(s)}$

Explanation of Solution

The balanced chemical reaction equation for any reaction is written in accordance with the law of conservation of mass.

Here iron metal is reacting with chlorine gas to give iron (III) chloride.

The balanced chemical reaction equation for this reaction is,

  $2F{e}_{(s)}+3C{l}_2{}_{(g)}\stackrel{}{\to }2FeC{l}_3{}_{(s)}$

Interpretation Introduction

Interpretation:

The mass of chlorine required to complete the reaction with $\text{10}\text{.0}\text{g}$ of $\text{Fe}$ has to be determined.  Mass of ${\text{FeCl}}_{\text{3}}$ produced in the reaction has to be calculated.

Concept Introduction:

• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol).
• Product of a chemical reaction is the produced compounds or the compounds formed after a chemical reaction.
• Stoichiometric factor is a relationship between reactant and product which is obtained from the balanced chemical equation for a particular reaction.

Answer to Problem 82GQ

$\text{19}\text{.143}\text{g}{\text{Cl}}_{\text{2}}$ is required to react with $\text{10}\text{.0 g}$ of iron.

Mass of $FeC{l}_3$ produced in the reaction is $\text{29}\text{.196}\text{g}$.

Explanation of Solution

The balanced chemical reaction equation for this reaction is,

  $2F{e}_{(s)}+3C{l}_2{}_{(g)}\stackrel{}{\to }2FeC{l}_3{}_{(s)}$

The amount (in moles) of iron available can be calculated by using the equation,

  $\text{Number}\text{of}\text{mole}\text{=}\frac{\text{Given}\text{mass}\text{of}\text{the}\text{substance}}{\text{Molar}\text{mass}}$

  $\text{10}\text{.0}\text{g}\text{Fe}\text{×}\frac{\text{1}\text{mol}\text{Fe}}{\text{55}\text{.85}\text{g}\text{Fe}}=\text{0}\text{.18}\text{mol}\text{Fe}$

Using the stoichiometric factor calculating the amount of chlorine required based on the amount of iron as follows,

  $\text{0}\text{.18}\text{mol}\text{Fe}\text{×}\frac{\text{3}\text{mol}{\text{Cl}}_{\text{2}}}{\text{2}\text{mol}\text{Fe}}\text{=}\text{0}\text{.27}\text{mol}{\text{Cl}}_{\text{2}}$

$\begin{array}{l}{\text{The mass of Cl}}_{\text{2}}\text{ required}\text{= }\text{0}\text{.27}\text{mol}{\text{Cl}}_{\text{2}}\text{×}\frac{\text{70}\text{.9}\text{g}{\text{Cl}}_{\text{2}}}{\text{1}\text{mol}{\text{Cl}}_{\text{2}}}\\ \text{=}\text{19}\text{.143}\text{g}{\text{Cl}}_{\text{2}}\end{array}$

From the balanced equation it is clear that the ratio between $FeandFeC{l}_3$ is $1:1$

Therefore the amount of $FeC{l}_3$ is $\text{0}\text{.18}\text{mol}$.

Mass of $FeC{l}_3$ can be determined by multiplying the amount of $FeC{l}_3$ with its molar mass.

Thus,

Mass of $FeC{l}_3$ produced $\text{=}\text{0}\text{.18}\text{mol}\text{×}\text{162}\text{.2}\text{g/mol}\text{=}\text{29}\text{.196}\text{g}$

Interpretation Introduction

Interpretation:

The percent yield of $FeC{l}_3$ has to be determined when $\text{18}\text{.5 g}$ of ${\text{FeCl}}_{\text{3}}$ is obtained from $\text{10}\text{.0 g}$ of iron and excess of ${\text{Cl}}_{\text{2}}\text{.}$

Concept Introduction:

• $Percentageyield=\frac{Actualyield}{Theoreticalyield}\times 100$
• Theoretical yield is the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry.

Answer to Problem 82GQ

The percent yield of $FeC{l}_3$ is $63.4\%.$

Explanation of Solution

If only $18.5g$ of $FeC{l}_3$ is obtained from $10.0g$ of iron and excess $C{l}_2$ the percent yield of $FeC{l}_3$ can be calculated by using the below mentioned equation,

  $\begin{array}{l}PercentageyieldofFeC{l}_3=\frac{ActualyieldofFeC{l}_3}{TheoreticalyieldofFeC{l}_3}\times 100\\ =\frac{18.5g}{29.196g}\times 100\\ =63.4\%\end{array}$

Interpretation Introduction

Interpretation:

The theoretical yield of $FeC{l}_3$ when $10.0g$ of iron and $10.0g$ of chlorine combined has to be determined.

Concept Introduction:

Limiting reagent: limiting reagent is a reactant, which consumes completely in the chemical reaction. The quantity of the product depends on this limiting reagent, it can be determined with the help of balanced chemical equation for the reaction.

Answer to Problem 82GQ

The theoretical yield of $FeC{l}_3$ when $10.0g$ of each of iron and chlorine combined is $\text{14}\text{.6}\text{g}$

Explanation of Solution

If $10.0g$ of each of iron and chlorine combined to produce $FeC{l}_3$, chlorine will acts as the limiting reagent.

So the maximum product yield that can be expected based on the masses of the reactants and the reaction stoichiometry is $\text{14}\text{.6}\text{g}$

Therefore the theoretical yield of $FeC{l}_3$ when $10.0g$ of each of iron and chlorine combined is $\text{14}\text{.6}\text{g}$