Chapter 6, Problem 6.133QE

Interpretation Introduction

Interpretation:

Pressure in $10\text{ L}$ container filled with $2.66{\text{ g H}}_2$ and $\text{4}{\text{.88 g Cl}}_2$ that is heated upto $111°\text{C}$ has to be calculated.

Concept Introduction:

Any gas that obeys the assumption laid down in kinetic molecular theory is said to be an ideal gas. The combination of all the gas laws namely Boyle’s law, Charles law and Avogadro’s leads to the ideal gas equation that can be used to relate all the four properties such as given below:

  $PV=nRT$

Here,

$R$ is the universal gas constant.

$V$ denotes the volume.

$n$ denotes the number of moles.

$T$ denotes temperature.

$P$ denotes the pressure.

Answer to Problem 6.133QE

Pressure in $10\text{ L}$ container filled with $2.66{\text{ g H}}_2$ and $\text{4}{\text{.88 g Cl}}_2$ is $4.38\text{ atm}$.

Explanation of Solution

The formula to convert degree Celsius to kelvin is as follows:

  $T\left(\text{K}\right)=T\left(°\text{C}\right)+273\text{ K}$        (1)

Here,

$T\left(\text{K}\right)$ denotes the temperature in kelvins.

$T\left(°\text{C}\right)$ denotes the temperature in Celsius.

Substitute $\text{111 }°\text{C}$ for $T\left(°\text{C}\right)$ in equation (1).

  $\begin{array}{c}T\left(\text{K}\right)=\text{111 }°\text{C}+273\text{ K}\\ =384\text{ K}\end{array}$

The formula to convert mass in gram to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$        (2)

Substitute $\text{2}\text{.66 g}$ for mass and $\text{2}\text{.0158 g/mol}$ for molar mass to calculate the number of moles of ${\text{H}}_2$ in equation (2).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{2}\text{.66 g}}{\text{2}\text{.0158 g}}\\ =1.31957\text{ mol}\end{array}$

Substitute $\text{4}\text{.88 g}$ for mass and $\text{70}\text{.906 g/mol}$ for molar mass to calculate the number of moles of ${\text{Cl}}_2$ in equation (2).

  $\begin{array}{c}\text{Number of moles}=\frac{\text{4}\text{.88 g}}{\text{70}\text{.906 g/mol}}\\ =0.068823\text{ mol}\end{array}$

The balanced chemical reaction when iron combines with $\text{HCl}$ is given as follows:

  ${\text{H}}_2+{\text{Cl}}_2\to 2\text{HCl}$        (3)

According to the stoichiometry of the balanced equation (3), one mole of ${\text{H}}_2$ react with one mole of ${\text{Cl}}_2$ to produce two moles of $\text{HCl}$, therefore number of moles of $\text{HCl}$ produced from $0.068823{\text{ mol Cl}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Number of moles of HCl}=\left(0.068823{\text{ mol Cl}}_2\right)\left(\frac{\text{2 mol HCl}}{1{\text{ mol Cl}}_2}\right)\\ =0.137647\text{ mol HCl}\end{array}$

Number of moles of $\text{HCl}$ produced from $1.31957{\text{ mol H}}_2$ is calculated as follows:

  $\begin{array}{c}\text{Number of moles of HCl}=\left(1.31957{\text{ mol H}}_2\right)\left(\frac{\text{2 mol HCl}}{1{\text{ mol H}}_2}\right)\\ =2.63914\text{ mol HCl}\end{array}$

Since ${\text{Cl}}_2$ yields lesser number of moles so it must be the limiting reagent, hence the number of moles of $\text{HCl}$ obtained from this reaction is $0.137646\text{ mol}$.

Out of the total, $1.31957{\text{ mol H}}_2$ the amount of ${\text{H}}_2$ left after it has reacted with $0.068823{\text{ mol Cl}}_2$ is calculated as flows:

$\begin{array}{c}{\text{Number of moles of excess H}}_{\text{2}}=1.31957\text{ mol}-0.068823\text{ mol}\\ \text{=1}\text{.250747 mol}\end{array}$

Rearrange equation (1) to obtain the expression of temperature as follows:

  $P=\frac{n_{\text{F}}RT}{V}$        (4)

The value of $n_F$ is calculated as follows:

  $n_{\text{F}}=n_{{\text{H}}_2}+n_{\text{HCl}}$        (5)

Substitute $\text{1}\text{.250747 mol}$ for $n_{{\text{Cl}}_2}$ and $0.137647\text{ mol}$  for $n_{\text{HCl}}$ in equation(5).

  $\begin{array}{c}{n}_F=\text{1}\text{.250747 mol}+0.137647\text{ mol}\\ =\text{1}\text{.3883 mol}\end{array}$

Substitute $\text{1}\text{.3883 mol}$ for $n_F$, $\text{10 L}$ for $V$,and $0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}$ for $R$ and $384\text{ K}$ for $T$ in equation (4).

  $\begin{array}{c}P=\frac{\left(\text{1}\text{.3883 mol}\right)\left(0.08206\text{ L}\cdot \text{atm/mol}\cdot \text{K}\right)\left(384\text{ K}\right)}{\left(\text{10 L}\right)}\\ =4.38\text{ atm}\end{array}$