Question

Calculate the mass of Potassium chlorate required to liberate $6.72d{m^3}$ of oxygen at $STP$ . Molar mass of $KCl{O_3}$ is $122.5gmo{l^{ - 1}}$

Answer 1

The given question can be solved by considering, the volume of one mole of any gas is known as molar volume is equal to $22.4L$ at standard temperature and pressure. Molar volume allows conversion to be made between volume and moles of any gases at standard temperature.

Complete step by step answer:
In the question it is given that the volume of oxygen liberated is equal to $6.72d{m^3}$ . So, one mole of oxygen,
$1$ mole of oxygen$= 22.4d{m^3} = 1$ mole
Now, $6.72d{m^3}$ contain $\dfrac{{6.72}}{{22.4}}$ moles of oxygen gas.
$6.72d{m^3}$ of oxygen contains $0.3$ moles of oxygen gas. Thus, the reaction can be written as
$2KCl{O_3}\left( s \right)\xrightarrow{\Delta }2KCl\left( s \right) + 3{O_2}$
From the reaction it may conclude that $2mol$ of $KCl{O_3}$ gives $2mol$ of $KCl$ and $3mol$ of oxygen gas. It means that if we liberate $3mol$ of oxygen gas, we require $2mol$ of $KCl{O_3}$ .
So, $1$ mole of oxygen will require $\dfrac{2}{3}$moles of $KCl{O_3}$.
Now, from above statement $1$ mole of oxygen requires $0.66$ moles of $KCl{O_3}$ .
So, for $0.3$ of oxygen it will require $0.66 \times 0.3 = 0.2mol$ of $KCl{O_3}$ .
We have molar mass and moles of $KCl{O_3}$ , we can calculate the mass of potassium chlorate. By using formulas which relate mass, molar mass and moles of $KCl{O_3}$ .
We know the number of moles of a substance is equal to the mass of the substance divided by its molar mass.
By putting the value of molar mass and number of moles we can calculate the mass of $KCl{O_3}$ .
$mass = 0.2 \times 122.5$
$mass = 24.5g$

Note:
It is to be noted that one mole of any gas contains $22.4L$ volume at standard temperature and pressure of gas. We can calculate the volume at $STP$ by using the ideal gas equation. The ideal gas reaction is $PV = nRT$ .