Chapter 20, Problem 20.79QE

Interpretation Introduction

Interpretation:

For the preparation of Oxygen from ${\text{KClO}}_{\text{3}}$ reaction equation has to be written, and also the mass of ${\text{KClO}}_{\text{3}}$ required to produce $\text{0}\text{.50}\text{L}$ Oxygen has to be determined.

Concept Introduction:

Ideal gas equation:

The ideal gas equation is given by,

  $\text{P}\text{V}\text{=}\text{n}\text{R}\text{T}$

Where,

  $\text{P}$ is the pressure,

  $\text{V}$ is the volume,

  $\text{T}$ is the temperature,

  $\text{R}$ is molar gas constant,

  $n$ is the number of moles.

Explanation of Solution

Given information,

  Volume of Oxygen is $\text{0}\text{.50}\text{L}$, temperature is $\text{27}{}^{\text{o}}\text{C}$ and pressure is $\text{755}\text{torr}$.

Preparation of Oxygen from ${\text{KClO}}_{\text{3}}$,

  $\text{2}{\text{KClO}}_{\text{3}\left(\text{s}\right)}\stackrel{{\text{MnO}}_{\text{2}}}{\to }\text{2}{\text{KCl}}_{\left(\text{s}\right)}\text{+}\text{3}{\text{O}}_{\text{2}\left(\text{g}\right)}$

In the above reaction, 2 moles of ${\text{KClO}}_{\text{3}}$ gives 3 moles of Oxygen.  So, the mole ratio between ${\text{KClO}}_{\text{3}}$ and Oxygen is $\text{2}\text{:}\text{3}$.

Calculate the moles of Oxygen formed in the above reaction by using ideal gas equation,

  $\begin{array}{l}\text{P}\text{V}\text{=}\text{n}\text{R}\text{T}\\ \\ \text{here}\text{the}\text{values}\text{of}\\ \text{P}\text{=}\text{755}\text{torr,}\text{V}\text{=}\text{0}\text{.50}\text{L,}\text{R}\text{=62}\text{.36}\text{L}\text{.torr/K}\text{.mol}\text{and}\text{T}\text{=}\text{300}\text{K}\\ \\ \text{n}\text{=}\frac{\text{P}\text{V}}{\text{R}\text{T}}\text{=}\frac{\text{755}\text{torr}\text{×}\text{0}\text{.50}\text{L}}{\left(\text{62}\text{.36}\text{L}\text{.torr/K}\text{.mol}\right)\text{×}\text{300}\text{K}}\\ \\ \text{=}\frac{\text{377}\text{.5}}{\text{18708}}\text{mol}\text{=}\text{0}\text{.02018}\text{mol}\end{array}$

Calculate the required mass of ${\text{KClO}}_{\text{3}}$ using the moles of Oxygen,

  $\begin{array}{l}\text{Mass}\text{=}\text{Moles}\text{×}\text{Molar}\text{mass}\\ \\ \text{Molar}\text{mass}\text{of}{\text{KClO}}_{\text{3}}\text{=}\text{122}\text{.55}\text{g}\text{/}\text{mol}\\ \\ \text{Mass}\text{of}{\text{KClO}}_{\text{3}}\text{=}\frac{\text{2}}{\text{3}}\left(\text{0}\text{.02018}\text{mol}\text{×}\text{122}\text{.55}\text{g/mol}\right)\\ \\ \text{=}\frac{\text{2}}{\text{3}}\left(\text{2}\text{.4731}\right)\text{g}\text{=}\text{1}\text{.6}\text{g}\end{array}$

Therefore, the required mass of ${\text{KClO}}_{\text{3}}$ is $\text{1}\text{.6}\text{g}$.