Question

Solid potassium chlorate (KClO3) decomposes into potassium chloride and oxygen gas when heated. How many moles of oxygen form when 55.3 g completely decomposes?

Answer 1

Concepts and reason

The concepts used here to calculate the moles of oxygen is the mole concept. A mole provides the specific measure of a number of atom or molecule in a bulk sample of matter.

Fundamentals

A number of the mole is defined as a given mass of solute to the molar mass of solute. It is expressed in (mol).

$n{\rm{ }} = {\rm{ }}\frac{m}{M}$ …… (1)

Here, $m$ is given mass and $M$ is molar mass.

The number of moles of ${\rm{(KCl}}{{\rm{O}}_3})$ is calculated by substituting the values of $M$ as ${\rm{122}}{\rm{.55 g mo}}{{\rm{l}}^{ - 1}}$ and $m$ as ${\rm{55}}{\rm{.3 g}}$ in the equation (1).

$\begin{array}{c}\\n{\rm{ = }}\frac{{{\rm{55}}{\rm{.3 }}\cancel{{\rm{g}}}}}{{{\rm{122}}{\rm{.55 }}\cancel{{\rm{g}}}{\rm{ mo}}{{\rm{l}}^{{\rm{ - 1}}}}}}\\\\n = 0.45{\rm{ mol}}\\\end{array}$

The balanced chemical equation of decomposition of potassium chlorate ${\rm{(KCl}}{{\rm{O}}_3})$ is shown below.

$\begin{array}{l}\\{\rm{2KCl}}{{\rm{O}}_3}\longrightarrow{\Delta }{\rm{ 2KCl}} + {\rm{ 3}}{{\rm{O}}_{\rm{2}}}\\\\{\rm{2 mol 3 mol }}\\\\{\rm{1 mol }}\frac{3}{2}{\rm{ mol }}\\\end{array}$

Now, calculating the number of moles of oxygen:

$\begin{array}{c}\\{\rm{2 moles of KCl}}{{\rm{O}}_3} \to 3{\rm{ moles of }}{{\rm{O}}_{\rm{2}}}\\\\{\rm{1 mole of KCl}}{{\rm{O}}_3} \to \frac{3}{2}{\rm{ moles of }}{{\rm{O}}_{\rm{2}}}\\\\{\rm{0}}{\rm{.45 moles of KCl}}{{\rm{O}}_3} \to \frac{3}{2} \times {\rm{0}}{\rm{.45 moles of }}{{\rm{O}}_{\rm{2}}}\\\\{\rm{ 0}}{\rm{.675 moles of }}{{\rm{O}}_2}\\\end{array}$

Ans:

The number of moles of oxygen formed on the decomposition of potassium chlorate are ${\bf{0}}{\bf{.675 mol}}$ .