Question

When heated, potassium permanganate decomposes according to the following equation:
$2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}$
Given that the molecular mass of potassium permanganate is 158, what volume of oxygen (measured at room temperature) would be obtained by the complete decomposition of 15.8 g of potassium permanganate?(Molar volume at room temperature is 24 litres.

Answer 1

Hint: For the solution of the above equation we need to understand the basics of mole concept. We can define that a mole of a substance or a mole of particles as exactly $6.02214076{\text{ }} \times {10^{23}}$ particles, which may be atoms, molecules, ions, or electrons. It is the unit of measurement for the amount of substance in SI units. With the help of the mole concept we can further calculate the molar mass of any substance.

Complete step by step answer:
As we know that one mole is equal to $6.022{\text{ }} \times {10^{23}}$molecular entities (Avogadro's number), and each element has a different molar mass depending on the weight of $6.022{\text{ }} \times {10^{23}}$ its atoms. Then we can determine the molar mass of any element by finding out the atomic mass of the element present on the periodic table.
Now, to solve the question let’s first write down the molecular mass of all the elements:
\[K = 39\],\[Mn = 55\] , \[O = 16\]
We know that, when potassium permanganate is heated then the equation becomes $2KMn{O_4} \to {K_2}Mn{O_4} + Mn{O_2} + {O_2}$
First, we will write down all the given quantities:
So, molecular mass of ${K_2}Mn{O_4}$ = \[158{\text{ }}g\]
Molar volume of ${O_2}$ at room temperature = 24 L
Now, we will substitute these values in place of the above equation:
Thus, \[2\; \times 158{\text{ }}g\]of ${K_2}Mn{O_4}$at the room temperature yields 24 L of ${O_2}$
So, using unitary method:
For 15.8 g of ${K_2}Mn{O_4}$at the room temperature yields = $\dfrac{{24 \times 15.8}}{{2 \times 158}}$ = $\dfrac{{379.2}}{{316}}$ = 1.2 g
$\therefore $\[1.2{\text{ }}g\] of ${O_2}$ would be obtained by the complete decomposition.

Note:
We must know that the Avogadro's number is a proportion that relates molar mass on an atomic scale to physical mass. And we know the avogadro’s number is equal to $6.022{\text{ }} \times {10^{23}}$moles per litre.