Chapter 5, Problem 152CP

You have an equimolar mixture of the gases SO₂ and O₂, along with some He, in a container fitted with a piston. The density of this mixture at STP is 1.924 g/L. Assume ideal behavior and constant temperature and pressure.

a. What is the mole fraction of He in the original mixture?

b. The SO₂ and O₂ react to completion to form SO₃. What is the density of the gas mixture after the reaction is complete?

Interpretation Introduction

Interpretation: The mole fraction of $\text{He}$ in the original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ should be determined, if the density of original mixture is $\text{1}\text{.924}\text{g/L}$ at constant temperature and pressure.

Concept introduction:

• Equation for density and number of moles are,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$  and  $\text{Number of moles}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Therefore,

According to ideal gas equation for molar mass in terms of density is,

$\text{Molar}\text{mass}\text{=}\frac{\text{Density}\text{×}\text{R}\text{×}\text{Temperature}}{\text{pressure}}$

• Mass of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{taken}\text{Mass}$

• Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

• $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

• Balanced equation of a reaction is written according to law of conservation of mass.

• 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.

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

To determine: The mole fraction of $\text{He}$ in the original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ and the density of gas mixture after completion of given reaction between ${\text{SO}}_2$ and ${\text{O}}_2$

Answer to Problem 152CP

The mole fraction of $\text{He}$ in the original mixture is $0.1114$.

Explanation of Solution

To find: the total molar mass of given original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$.

The total molar mass of given original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ is $\text{43}\text{.13}\text{g}$.

The given data’s of the given original mixture are,

The density of given mixture is given as $\text{1}\text{.924}\text{g/L}$.

Temperature $\text{= 273}\text{.2 K}$

Pressure $\text{=1atm}$

According to ideal gas equation for molar mass in terms of density is,

$\text{Molar}\text{mass}\text{=}\frac{\text{Density}\text{×}\text{R}\text{×}\text{Temperature}}{\text{pressure}}$

Therefore, the molar mass of given original mixture is,

$\begin{array}{l}\text{Molar}\text{mass}\text{=}\frac{\text{1}\text{.924}\text{g/L}\text{×}\text{0}\text{.08206}\text{L}\text{atm/}\text{K}\text{mol}\text{×}\text{273}\text{.2}\text{K}}{\text{1}\text{atm}}\\ \text{=}\text{43}\text{.13}\text{g/mol}\\ \end{array}$

To find: the number of moles of ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ in the original mixture.

Assumes the total number of moles in the original mixture is 1 mole.

The number of moles of ${\text{SO}}_2$ in the original mixture is $\text{0}\text{.4443}\text{mol}$.

The number of moles of ${\text{O}}_2$ in the original mixture is $\text{0}\text{.4443}\text{mol}$.

The number of moles of $\text{He}$ in the original mixture is $\text{0}\text{.1114}\text{mol}$.

Let’s take the number of moles of ${\text{SO}}_2$ and ${\text{O}}_2$ are x, since the both are equimolar and the number of moles of $\text{He}$ is z.

The total number of moles is 1 mole.

Therefore,

${\text{n}}_{{\text{SO}}_{\text{2}}}\text{+}{\text{n}}_{{\text{O}}_{\text{2}}}\text{+}\text{}\text{}{\text{n}}_{\text{He}}\text{=}\text{1}\text{mol}\text{Equation\_1}$

$\text{x}\text{+}\text{x}\text{+}\text{}\text{}\text{z}\text{=}\text{1}\text{mol}\text{Equation\_1}$

The total molar mass of given original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ is calculated as $\text{43}\text{.13}\text{g}$.

The molar mass of ${\text{SO}}_2$, ${\text{O}}_2$ and $\text{He}$ are 64g, 32g and 4g respectively.

The Mass of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{taken}\text{Mass}$

Therefore,

$\text{64}\text{x}\text{+}\text{32}\text{x}\text{+}\text{4}\text{z}\text{=}\text{43}\text{.13}\text{Equation\_2}$

To solve equation 1 and 2, subtract the equation _2 by equation_1multiplied with 4,

$\begin{array}{l}\left[\text{64}\text{.07}\text{x}\text{+}\text{32}\text{x}\text{+}\text{4}\text{z}\text{=}\text{43}\text{.13}\right]\text{-}\text{4}\left[\text{x}\text{+}\text{x}\text{+}\text{}\text{}\text{z}\text{=}\text{1}\right]\\ =\text{88}\text{.07}\text{x}\text{=}\text{}\text{39}\text{.13}\\ \text{x}\text{=}\text{0}\text{.4443}\end{array}$

Therefore, the value of z is $1-0.4443+0.4443=0.1114$

Hence,

The number of moles of ${\text{SO}}_2$ in the original mixture is $\text{0}\text{.4443}\text{mol}$.

The number of moles of ${\text{O}}_2$ in the original mixture is $\text{0}\text{.4443}\text{mol}$.

The number of moles of $\text{He}$ in the original mixture is $\text{0}\text{.1114}\text{mol}$.

To determine: The mole fraction of $\text{He}$ in the original mixture containing ${\text{SO}}_2$, ${\text{O}}_2$.

The mole fraction of $\text{He}$ in the original mixture is $0.1114$.

The number of moles of $\text{He}$ in the original mixture is $\text{0}\text{.1114}\text{mol}$.

The total number of moles in original mixture is 1 mole.

Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

$\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

Therefore,

The mole fraction of $\text{He}$ in the mixture is,

$\frac{\text{0}\text{.1114}\text{mol}}{\text{1}\text{mol}}=\text{0}\text{.1114}$

Hence,

The mole fraction of$\text{He}$ in the original mixture is $0.1114$.

Interpretation Introduction

Interpretation: The density of gas mixture after completion of given reaction between ${\text{SO}}_2$ and ${\text{O}}_2$ should be determined, if the density of original mixture is $\text{1}\text{.924}\text{g/L}$ at constant temperature and pressure.

Concept introduction:

• Equation for density and number of moles are,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$  and  $\text{Number of moles}\text{=}\frac{\text{Mass}}{\text{Molar}\text{mass}}$

Therefore,

According to ideal gas equation for molar mass in terms of density is,

$\text{Molar}\text{mass}\text{=}\frac{\text{Density}\text{×}\text{R}\text{×}\text{Temperature}}{\text{pressure}}$

• Mass of a substance from its number of moles is,

$\text{Number of moles}\text{×}\text{Molecular}\text{mass in grams}\text{=}\text{taken}\text{Mass}$

• Equation for mole fraction of a molecule in a mixture of three molecules (A, B and C) is,

• $\text{molecule}\text{fraction}\text{of}\text{A,}\text{(}{\text{χ}}_{\text{A}}\text{) }\text{=}\frac{\text{numbers}\text{of moles of}\text{molecule A}{\text{(n}}_{\text{A}}\text{)}}{\text{total number of moles}{\text{(n}}_{\text{A}}{\text{+n}}_{\text{B}}{\text{+n}}_{\text{C}}\text{) }}$

• Balanced equation of a reaction is written according to law of conservation of mass.

• 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.

• Mole ratios between the reactant and products of a reaction are depends upon the coefficients of respective reactant in a balanced chemical equation.

• Partial pressure of a gas in a mixture of gases is the pressure of that gas when it alone.

To determine: The density of gas mixture after completion of given reaction between ${\text{SO}}_2$ and ${\text{O}}_2$

Answer to Problem 152CP

The density of gas mixture after completion of given reaction is $\text{2}\text{.473}\text{g/L}$.

Explanation of Solution

To find: the balanced equations for given reaction.

${\text{2SO}}_2\text{+}{\text{O}}_{\text{2}}\to 2{\text{SO}}_3$

The reactants presented are ${\text{SO}}_2$, ${\text{O}}_2$.

The product of the given reactions is ${\text{SO}}_3$.

Therefore,

The chemical equations for the given reactions are,

${\text{SO}}_2\text{+}{\text{O}}_{\text{2}}\to {\text{SO}}_3$

This equation is not balanced, Balanced equation is the chemical equation of a reaction which is written according to law of conservation of mass the mass of reactant molecules should conserve in product molecules.

The balanced chemical equation of the given reaction is,

Balance the molecules in both sides of arrow in order of atoms other than oxygen atoms and then oxygen atoms.

Therefore, the balanced chemical equation of the given reaction will be,

${\text{2SO}}_2\text{+}{\text{O}}_{\text{2}}\to 2{\text{SO}}_3$

Hence,

The balanced chemical equation for the given reaction is, ${\text{2SO}}_2\text{+}{\text{O}}_{\text{2}}\to 2{\text{SO}}_3$.

To find: the limiting reagent for given reaction.

The limiting reagent for given reaction is ${\text{SO}}_2$.

The balanced chemical equation for the given reaction is, ${\text{2SO}}_2\text{+}{\text{O}}_{\text{2}}\to 2{\text{SO}}_3$.

2 mole of ${\text{SO}}_2$ is reacting with 1 mole of ${\text{O}}_2$. Since both are taken as equimolar in mixture, the ${\text{SO}}_2$ will finish first in given reaction.

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.

Therefore, here in this reaction ${\text{SO}}_2$ is the limiting reagent.

To find: the number of moles of mixture after the reaction.

#assume the total number of moles is 1 mole.

The number of moles of mixture after the reaction is $\text{0}\text{.7779}\text{mol}$.

The final mixture contains product ${\text{SO}}_3$ and remaining ${\text{O}}_2$.

The balanced chemical equation for the given reaction is, ${\text{2SO}}_2\text{+}{\text{O}}_{\text{2}}\to 2{\text{SO}}_3$.

The limiting reagent for given reaction is ${\text{SO}}_2$ and the number of moles of ${\text{SO}}_2$ is $\text{0}\text{.4443}\text{mol}$.

Therefore,

The number of moles of ${\text{SO}}_3$ is,

$\text{0}\text{.4443}\text{mol}{\text{SO}}_{\text{2}}\text{×}\frac{\text{2}\text{mol}{\text{SO}}_{\text{3}}}{\text{2}\text{mol}{\text{SO}}_{\text{2}}}\text{=}\text{0}\text{.4443}\text{mol}{\text{SO}}_{\text{3}}$

The number of moles of reacting ${\text{O}}_2$ is,

$\text{0}\text{.4443}\text{mol}{\text{SO}}_{\text{2}}\text{×}\frac{1\text{mol}{\text{O}}_2}{\text{2}\text{mol}{\text{SO}}_{\text{2}}}\text{=}\text{0}\text{.2222}\text{mol}{\text{O}}_2$

The number of moles of ${\text{O}}_2$ in the original mixture is $\text{0}\text{.4443}\text{mol}$.

Therefore, the number of moles of reacting ${\text{O}}_2$ is $\text{0}\text{.2222}\text{mol}$.

Then, the number of moles of final mixture containing product ${\text{SO}}_3$ and remaining ${\text{O}}_2$ is,

$\text{0}\text{.4443}\text{mol}\text{+}\text{0}\text{.2222}\text{mol}\text{=}\text{0}\text{.7779}\text{mol}$

Hence,

The number of moles of mixture after the reaction is $\text{0}\text{.7779}\text{mol}$.

To determine: the density of gas mixture after completion of given reaction between ${\text{SO}}_2$ and ${\text{O}}_2$

The density of gas mixture after completion of given reaction is $\text{2}\text{.473}\text{g/L}$.

The density of given initial mixture is given as $\text{1}\text{.924}\text{g/L}$.

The total number of moles in initial mixture is 1 mole.

The number of moles of mixture after the reaction is $\text{0}\text{.7779}\text{mol}$.

Equation for density is,

$\text{Density}\text{=}\frac{\text{Mass}}{\text{Volume}}$

Volume of gas, according to ideal gas equation,

$\text{Volume}\text{=}\frac{\text{Number}\text{of}\text{moles}\times \text{R}\text{×}\text{Temperature}}{\text{Pressure}}$

The reaction is at constant temperature and pressure.

Therefore,

$\left\{\begin{array}{l}\text{density}\propto \frac{\text{1}}{\text{volume}}\\ \text{volume}\propto \text{number}\text{of}\text{moles}\end{array}\right\}\Rightarrow \text{density}\propto \frac{1}{\text{number}\text{of}\text{moles}}$

That means,

$\frac{{\text{d}}_{\text{initial}}}{{\text{d}}_{\text{final}}}\text{=}\frac{{\text{n}}_{\text{final}}}{{\text{n}}_{\text{initial}}}$

Therefore,

$\begin{array}{l}\frac{\text{1}\text{.924}\text{g/L}}{{\text{d}}_{\text{final}}}\text{=}\frac{\text{0}\text{.7779}\text{mol}}{\text{1}\text{mol}}\\ {\text{d}}_{\text{final}}\text{=}\text{2}\text{.473}\text{g/L}\end{array}$

Hence,

The density of gas mixture after completion of given reaction is $\text{2}\text{.473}\text{g/L}$.