Chapter 10, Problem 106IL

(a)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the balanced equation, moles, mole fraction of the reactant, the unreacted amount of given reactant, partial pressure of given compounds should be determined.

Concept Introduction:

Balanced Chemical Equation:

The chemical reaction when the number of atoms present in the reactant side of the reaction should be equal to the number and the charge of atoms present in the product side of the reaction which then only be considered as balanced.

Answer to Problem 106IL

  ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}{\text{+5O}}_{\text{2}\left(\text{g}\right)}\to 3{\text{CO}}_{\text{2}\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(l\right)}$

Explanation of Solution

The given reactants ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}$ and ${\text{O}}_{\text{2}\left(\text{g}\right)}$ reacts in order to give set of products like ${\text{CO}}_{\text{2}\left(\text{g}\right)}$ and ${\text{H}}_{\text{2}}{\text{O}}_{\left(l\right)}$. In order to obtain the balanced chemical reaction the number of atoms present in given reactant and products are analyzed and the suitable coefficients are included before them in order to obtain the balanced equation.

Observing given reactant and products 3 should be included before ${\text{CO}}_{\text{2}\left(\text{g}\right)}$ in order to get equal number of carbon atoms on both sides then, 4 should be included before ${\text{H}}_{\text{2}}{\text{O}}_{\left(l\right)}$ in order to obtain equal number of hydrogen atom on both sides and finally, 5 should be included before ${\text{O}}_{\text{2}\left(\text{g}\right)}$ so that the number of oxygen atoms present are balanced on the both sides of the equation.

Therefore, the balanced chemical equation is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}{\text{+5O}}_{\text{2}\left(\text{g}\right)}\to 3{\text{CO}}_{\text{2}\left(\text{g}\right)}{\text{+4H}}_{\text{2}}{\text{O}}_{\left(l\right)}$.

(b)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the moles, of given compounds should be determined.

Concept introduction:

Ideal gas equation:

Any gas is described by using four terms namely pressure, volume, temperature and the amount of gas.  Thus combining three laws namely Boyle’s, Charles’s Law and Avogadro’s Hypothesis the following equation could be obtained.  It is referred as ideal gas equation.

  $\begin{array}{l}\text{V }\propto \frac{\text{nT}}{\text{P}}\\ \text{V = R}\frac{\text{nT}}{\text{P}}\\ \text{PV = nRT}\\ \text{where,}\\ \text{n = moles of gas}\\ \text{P = pressure}\\ \text{T = temperature}\\ \text{R = gas }\text{constant}\end{array}$

Under some conditions gases don not behave like ideal gas that is they deviate from their ideal gas properties.  At lower temperature and at high pressures the gas tends to deviate and behave like real gases.

Boyle’s Law:

At given constant temperature conditions the mass of given ideal gas in inversely proportional to its volume.

Charles’s Law:

At given constant pressure conditions the volume of ideal gas is directly proportional to the absolute temperature.

Avogadro’s Hypothesis:

Two equal volumes of gases with same temperature and pressure conditions tend to have same number of molecules with it.

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

    $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 106IL

The number of moles of given species is $\text{0}\text{.3178moles}$.

Explanation of Solution

The number of moles present in the given amount of the substance is calculated as follows,

  $\begin{array}{l}\text{PV = nRT}\\ \text{n = }\frac{\text{PV}}{\text{RT}}\\ \text{=}\frac{\text{5}\text{.1atm×1}\text{.50L}}{\text{0}{\text{.0821×20}}^{\text{o}}\text{C}}\text{=}\frac{\text{5}\text{.1atm×1}\text{.50L}}{\text{0}\text{.0821×}\left(\text{273}\text{.15+20}\right)}\\ \text{=}\frac{\text{7}\text{.65}}{\text{0}\text{.0821×293}\text{.15}}\text{=}\frac{\text{7}\text{.65}}{\text{24}\text{.07}}\text{=0}\text{.3178moles}\text{.}\\ \text{Since,}\\ {\text{P}}_{\text{total}}{\text{=Partialpressureof CO}}_{\text{2}\left(\text{g}\right)}{\text{+Partialpressureof H}}_{\text{2}}{\text{O}}_{\left(\text{l}\right)}\\ =0.10atm+5.0atm\\ =\text{5}\text{.1atm}\end{array}$

(c)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the balanced equation, moles, mole fraction of the reactant, the unreacted amount of given reactant, partial pressure of given compounds should be determined.

Concept introduction:

Mole fraction: The mole fraction of denotes the individual presence of the component present in the given chemical reaction.

Consider general equation that contains reactants X and Y then the mole fraction of X is determined as follows,

  $\begin{array}{l}\text{Mole fraction of Mole fraction of one component = }\frac{\text{Moles of that component}}{\text{Total moles present in the reaction}}\\ \text{Mole fraction of X = }\frac{\text{Number of moles of X}}{\text{Number of moles of X + Number of moles of Y}}\end{array}$

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

    $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 106IL

The mole fraction for the given species is equal to $0.1667$

Explanation of Solution

Consider the given chemical reaction and calculate the mole fraction of ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}$ as follows,

Before the reaction there are only reactants present in the flask.

$\begin{array}{l}{\text{Mole fraction of C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}\text{ = }\frac{{\text{Number of moles of C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}}{{\text{Number of moles of C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}{\text{ + Number of moles of O}}_{\text{2}\left(g\right)}}\\ =\frac{1}{1+5}=0.1667\end{array}$

(d)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the balanced equation, moles, mole fraction of the reactant, the unreacted amount of given reactant, partial pressure of given compounds should be determined.

Concept introduction:

Mole fraction: The mole fraction of denotes the individual presence of the component present in the given chemical reaction.

Consider general equation that contains reactants X and Y then the mole fraction of X is determined as follows,

  $\begin{array}{l}\text{Mole fraction of Mole fraction of one component = }\frac{\text{Moles of that component}}{\text{Total moles present in the reaction}}\\ \text{Mole fraction of X = }\frac{\text{Number of moles of X}}{\text{Number of moles of X + Number of moles of Y}}\end{array}$

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

    $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 106IL

The amount of unreacted oxygen in the reaction flask is equal to $\text{0}\text{.28045 moles}$.

Explanation of Solution

  $\begin{array}{l}\text{PV = nRT}\\ {\text{Initial moles, n}}_{{\text{O}}_{\text{2}}}\text{ = }\frac{PV}{RT}=\frac{5\times 1.50L}{0.0821\times 293.15K}=0.3116moles\end{array}$

Initial moles of oxygen is $0.3116moles$

  $\begin{array}{l}\text{PV = nRT}\\ {\text{Initial moles, n}}_{{\text{C}}_{\text{3}}{\text{H}}_{\text{8}}}\text{ = }\frac{PV}{RT}=\frac{0.1\times 1.50L}{0.0821\times 293.15K}=6.23\times {10}^{-3}moles\end{array}$

From the given conditions it is clear that ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}$ acts as limiting reagent hence the unreacted oxygen is calculated by subtracting the amount of available amount of oxygen with the amount needed by the ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}\left(\text{g}\right)}$.

  $\begin{array}{l}{\text{Amount of O}}_{\text{2}}{\text{ needed by C}}_{\text{3}}{\text{H}}_{\text{8}}\text{=}\text{6}{\text{.23×10}}^{\text{-3}}\text{×5}\text{=}\text{0}\text{.0315moles}\\ {\text{Excess of O}}_{\text{2}}\text{=}\text{0}\text{.3116-0}\text{.03115 = 0}{\text{.28045 moles of O}}_{\text{2}}\end{array}$

(e)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the of given compounds should be determined.

Concept introduction:

Mole fraction: The mole fraction of denotes the individual presence of the component present in the given chemical reaction.

Consider general equation that contains reactants X and Y then the mole fraction of X is determined as follows,

  $\begin{array}{l}\text{Mole fraction of Mole fraction of one component = }\frac{\text{Moles of that component}}{\text{Total moles present in the reaction}}\\ \text{Mole fraction of X = }\frac{\text{Number of moles of X}}{\text{Number of moles of X + Number of moles of Y}}\end{array}$

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

    $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 106IL

The partial pressure exerted by the given gas in the system is $\text{2}\text{.19 atm}$

Explanation of Solution

The partial pressure for the gas $C{O}_2$ is calculated as follows,

  $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \\ {\text{P}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{{\text{Moles of CO}}_{\text{2}}}{\text{Total moles}}{\text{×P}}_{\text{total}}\\ =\frac{3}{\text{7}}\text{×5}\text{.1 atm}\\ \text{=}\text{2}\text{.19 atm}\end{array}$

(f)

Interpretation Introduction

Interpretation:

For the given two reactants and the products under given set of conditions the partial pressure of given compounds should be determined.

Concept introduction:

Mole fraction: The mole fraction of denotes the individual presence exerted by the component present in the given chemical reaction.

Consider general equation that contains reactants X and Y then the mole fraction of X is determined as follows,

  $\begin{array}{l}\text{Mole fraction of Mole fraction of one component = }\frac{\text{Moles of that component}}{\text{Total moles present in the reaction}}\\ \text{Mole fraction of X = }\frac{\text{Number of moles of X}}{\text{Number of moles of X + Number of moles of Y}}\end{array}$

The relationship between partial pressure and ${\text{P}}_{\text{total}}$ is

    $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \text{where,}\\ {\text{P}}_{\text{i}}\text{=}\text{partial pressure}\\ {\text{χ}}_{\text{i}}\text{=}\text{molefraction}\\ {\text{P}}_{\text{total}}\text{=Totalpressure}\end{array}$

Answer to Problem 106IL

The partial pressure exerted by the excess of oxygen after the reaction is $\text{0}\text{.24 atm}$.

Explanation of Solution

The partial pressure for the unreacted oxygen is as follows,

  $\begin{array}{l}{\text{P}}_{\text{i}}\text{=}{\text{χ}}_{\text{i}}{\text{P}}_{\text{total}}\\ \\ {\text{P}}_{{\text{O}}_{\text{2}}}\text{=}\frac{{\text{Moles of O}}_{\text{2}}}{\text{Total moles}}{\text{×P}}_{\text{total}}\\ =\frac{0.28045}{6}\text{×5}\text{.1 atm}\\ \text{=}\text{0}\text{.24 atm}\end{array}$