Chapter 14, Problem 10PE

Interpretation Introduction

Interpretation:

The volume of oxygen at ${\text{27}}^{\text{ο}}\text{C}$ and $\text{733 torr}$ produced from the decomposition of $\text{5}{\text{.00×10}}^{\text{2}}\text{ g}$ of a $\text{3}\text{.0\%}$ ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is to be stated.

Concept Introduction:

The ideal gas law is the compilation of Charles’s Law $\left(\text{V}\propto \text{T}\right)$, Boyle’s Law $\left(\text{V}\propto \frac{\text{1}}{\text{P}}\right)$ and Avogadro’s Law $\left(\text{V}\propto \text{n}\right)$.

The ideal gas equation resultant of all the ideal gas law’s is shown below.

$\text{PV}=\text{nRT}$

Answer to Problem 10PE

The volume of oxygen at ${\text{27}}^{\text{ο}}\text{C}$ and $\text{733 torr}$ produced from the decomposition of $\text{5}{\text{.00×10}}^{\text{2}}\text{ g}$ of $\text{3}\text{.0\%}$ ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is $\text{5}\text{.6 L}$.

Explanation of Solution

The ideal gas equation is shown below.

$\text{PV}=\text{nRT}$ … (1)

Where,

• $\text{P}$ is the pressure.

• $\text{V}$ is the volume.

• $\text{n}$ is the number of moles.

• $\text{R}$ is the gas constant.

• $\text{T}$ is the temperature.

Rearrange the equation (1) to find out V.

$\text{V}=\frac{\text{nRT}}{\text{P}}$ … (2)

The reaction that occurs when ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ decomposes is given below.

${\text{2H}}_{\text{2}}{\text{O}}_{\text{2}}\left(\text{aq}\right)\to 2{\text{H}}_{\text{2}}\text{O}\left(\text{l}\right)+{\text{O}}_{\text{2}}\left(\text{g}\right)$

According to the above reaction, the number of moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reacted equals to the number of moles of water produced. Half of the number of moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$, oxygen gas are produced.

The total mass of a $\text{3}\text{.0\%}$ ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is $\text{5}{\text{.00×10}}^{\text{2}}\text{ g}$.

The mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reacted is calculated below.

$\begin{array}{rll}{\text{Mass of H}}_{\text{2}}{\text{O}}_{\text{2}}& =& 5.00\times {10}^2\text{g}\times \frac{3}{100}\\ & =& \text{15}\text{.0 g}\end{array}$

The conversion of mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ into the number of moles is shown below.

The formula to calculate the number of moles in terms of mass of element is shown below.

$\text{n}=\frac{\text{m}}{\text{M}}$ … (3)

Where,

• $\text{m}$ is the mass of gas.

• $\text{M}$ is the molar mass of gas.

The mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ reacted is $\text{15}\text{.0 g}$ and the molar mass of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $\text{34}{\text{.02 g mol}}^{-1}$.

Substitute the value $\text{m}$ and $\text{M}$ in equation (3).

$\begin{array}{rll}\text{n}& =& \frac{\text{m}}{\text{M}}\\ & =& \frac{\text{15}\text{.0 g}}{\text{34}{\text{.02 g mol}}^{-1}}\\ & =& \text{0}\text{.44 mol}\end{array}$

Number of moles of oxygen gas produced equals to the half of the number of moles of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ decomposed.

$\begin{array}{rll}\text{Number of moles of oxygen}& =& \frac{{\text{Number of moles of H}}_2{\text{O}}_{\text{2}}}{2}\\ & =& \frac{\text{0}\text{.44 mol}}{2}\\ & =& \text{0}\text{.22 mol}\end{array}$

The pressure and temperature are $\text{733 torr}$ and $\text{27}°\text{C}$ respectively.

The conversion of temperature in degree Celsius to Kelvin units is shown below.

$\text{T}\left(\text{K}\right)=\text{T}\left(°\text{C}\right)+273$ … (4)

Where,

• $\text{T}\left(\text{K}\right)$ is the temperature in Kelvin units.

• $\text{T}\left(°\text{C}\right)$ is the temperature in degree Celsius units.

Temperature of gas in degree Celsius units is $27°\text{C}$.

Substitute the value $\text{T}\left(°\text{C}\right)$ and in the equation 4.

$\begin{array}{rll}\text{T}\left(\text{K}\right)& =& \text{T}\left(°\text{C}\right)+273\\ & =& 27+\text{273}\\ & =& 300\text{K}\end{array}$

The conversion pressure from torr to atm is shown below.

$\begin{array}{rll}\text{760 torr}& =& \text{1 atm}\\ \text{1 torr}& =& \frac{\text{1 atm}}{\text{760 torr}}\\ \text{733 torr}& =& \frac{\text{1 atm}}{\text{760 torr}}\times \text{733 torr}\\ \text{733 torr}& =& \text{0}\text{.964 atm}\end{array}$

The value of gas constant is $0.0821\text{L}\text{atm}/\text{mol}\text{K}$.

Substitute the values of $\text{P}$, $\text{n}$, $\text{R}$ and $\text{T}$ in equation (2).

$\begin{array}{rll}\text{V}& =& \frac{\text{nRT}}{\text{P}}\\ & =& \frac{\text{0}\text{.22 mol}\times \left(0.0821\text{L}\text{atm}/\text{mol}\text{K}\right)\times 300\text{K}}{0.964\text{atm}}\\ & =& \frac{5.42}{\text{0}\text{.964}}\text{L}\\ & =& \text{5}\text{.6 L}\end{array}$

Therefore, the volume of oxygen at ${\text{27}}^{\text{ο}}\text{C}$ and $\text{733 torr}$ produced from the decomposition of $\text{5}{\text{.00×10}}^{\text{2}}\text{ g}$ of $\text{3}\text{.0\%}$ ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ solution is $\text{5}\text{.6 L}$.

Conclusion

The volume of oxygen at ${\text{27}}^{\text{ο}}\text{C}$ and $\text{733 torr}$ produced is $\text{5}\text{.6 L}$.