Chapter 3, Problem 3.8E

(a)

Interpretation Introduction

Interpretation:

Volume occupied by dry hydrogen has to be determined.

Concept Introduction:

Ideal gas law can be represented as equation for state for any imaginary gas. Expression for ideal gas equation is as follows:

  $PV=nRT$        (1)

Here,

$P$ is pressure of the gas.

$V$ is volume of gas.

$n$ denotes moles of gas.

$R$ is gas constant.

$T$ is temperature of gas.

Answer to Problem 3.8E

Volume of dry hydrogen is $0.128\text{ L}$.

Explanation of Solution

Balanced reaction for formation of ${\text{ZnCl}}_{\text{2}}$ is as follows:

  $\text{Zn}\left(\text{s}\right)+\text{2HCl}\left(\text{aq}\right)\to {\text{ZnCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Total external pressure must be equal to total internal pressure.

  $737.7\text{ Torr}={\text{Pressure of H}}_2+{\text{Vapor pressure of H}}_2\text{O}$        (2)

Rearrange equation (2) to determine pressure of ${\text{H}}_2$.

  ${\text{Pressure of H}}_2=737.7\text{ Torr}-{\text{Vapor pressure of H}}_2\text{O}$        (3)

Substitute $9.21\text{ Torr}$ for vapor pressure of ${\text{H}}_2\text{O}$ in equation (3).

  $\begin{array}{c}{\text{Pressure of H}}_2=737.7\text{ Torr}-9.21\text{ Torr}\\ =728.49\text{ Torr}\end{array}$

Conversion factor to convert $10°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =10°\text{C}+273.15\\ =283.15\text{K}\end{array}$

Rearrange equation (1) to determine value of $n$.

  $n=\frac{RT}{PV}$        (4)

Substitute $728.49\text{ Torr}$ for $P$, $127.0\text{ mL}$ for $V$, $62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $283.15\text{K}$ for $T$ in equation (4).

  $\begin{array}{c}n=\left(\frac{\left(728.49\text{ Torr}\right)\left(127\text{ mL}\right)}{\left(62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(283.15\text{K}\right)}\right)\left(\frac{{10}^{-3}\text{ L }}{1\text{ mL}}\right)\\ =\text{5}\text{.24}\times {\text{10}}^{-3}\text{ mol}\end{array}$

Rearrange equation (1) to determine value of $V$ for $\text{5}\text{.24}\times {\text{10}}^{-3}{\text{ mol H}}_2$.

  $V=\frac{nRT}{P}$        (5)

Substitute $1.0\text{ atm}$ for $P$, $\text{5}\text{.24}\times {\text{10}}^{-3}\text{ mol}$ for $n$, $8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $298\text{K}$ for $T$ in equation (5).

  $\begin{array}{c}V=\frac{\left(\text{5}\text{.24}\times {\text{10}}^{-4}\text{ mol}\right)\left(8.20574\times {10}^{-2}\text{ L}\cdot \text{atm}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(298\text{K}\right)}{1.0\text{ atm}}\\ =0.128\text{ L}\end{array}$

Hence, volume of dry hydrogen is $0.128\text{ L}$.

(b)

Interpretation Introduction

Interpretation:

Moles of dry hydrogen produced have to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3.8E

Moles of hydrogen produced is $\text{5}\text{.24}\times {\text{10}}^{-3}\text{ mol}$.

Explanation of Solution

Balanced reaction is as follows:

  $\text{Zn}\left(\text{s}\right)+\text{2HCl}\left(\text{aq}\right)\to {\text{ZnCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Total external pressure must be equal to total internal pressure.

  $737.7\text{ Torr}={\text{Pressure of H}}_2+{\text{Vapor pressure of H}}_2\text{O}$        (2)

Rearrange equation (2) to determine pressure of ${\text{H}}_2$.

  ${\text{Pressure of H}}_2=737.7\text{ Torr}-{\text{Vapor pressure of H}}_2\text{O}$        (3)

Substitute $9.21\text{ Torr}$ for vapor pressure of ${\text{H}}_2\text{O}$ in equation (3).

  $\begin{array}{c}{\text{Pressure of H}}_2=737.7\text{ Torr}-9.21\text{ Torr}\\ =728.49\text{ Torr}\end{array}$

Conversion factor to convert $10°\text{C}$ to Kelvin is as follows:

  $\begin{array}{c}\text{T}\left(\text{K}\right)=\text{temperature}\text{in}\text{celsius}+273.15\\ =10°\text{C}+273.15\\ =283.15\text{K}\end{array}$

Rearrange equation (1) to determine value of $n$.

  $n=\frac{RT}{PV}$        (4)

Substitute $728.49\text{ Torr}$ for $P$, $127.0\text{ mL}$ for $V$, $62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}$ for $R$, and $283.15\text{K}$ for $T$ in equation (4).

  $\begin{array}{c}n=\left(\frac{\left(728.49\text{ Torr}\right)\left(127\text{ mL}\right)}{\left(62.364\text{ L}\cdot \text{Torr}\cdot {\text{K}}^{-1}\cdot {\text{mol}}^{-1}\right)\left(283.15\text{K}\right)}\right)\left(\frac{{10}^{-3}\text{ L }}{1\text{ mL}}\right)\\ =\text{5}\text{.24}\times {\text{10}}^{-3}\text{ mol}\end{array}$

(c)

Interpretation Introduction

Interpretation:

Percentage purity of zinc has to be determined.

Concept Introduction:

Refer to part (a).

Answer to Problem 3.8E

Percentage purity of $\text{Zn}$ is $85.7\text{ \%}$.

Explanation of Solution

Balanced reaction is as follows:

  $\text{Zn}\left(\text{s}\right)+\text{2HCl}\left(\text{aq}\right)\to {\text{ZnCl}}_{\text{2}}\left(\text{aq}\right)+{\text{H}}_{\text{2}}\text{O}$

Number of moles of zinc present in $0.40\text{ g}$ can be calculated as follows:

  $\begin{array}{c}\text{Moles}=\frac{0.40\text{ g}}{ 65.38\text{ g}\cdot {\text{mol}}^{-1}}\\ =6.11\times {10}^{-3}\text{ mol}\end{array}$

Therefore, if zinc were pure then amount of hydrogen will be $6.11\times {10}^{-3}\text{ mol}$.

Percentage of purity of $\text{Zn}$ can be calculated as follows:

  $\begin{array}{c}\%\text{ Purity}=\left(\frac{\text{5}\text{.24}\times {\text{10}}^{-3}\text{ mol}}{6.11\times {10}^{-3}\text{ mol}}\right)\left(100\text{ \%}\right)\\ =85.7\text{ \%}\end{array}$

Hence percentage purity of $\text{Zn}$ is $85.7\text{ \%}$.