Question

${O_2}$ is evolved by heating $KCl{O_3}$ using $Mn{O_2}$ as a catalyst.
$2KCl{O_3}\xrightarrow{{Mn{O_2}}}2KCl + 3{O_2}$
Calculate the mass of $KCl{O_3}$ required to produce 6.72 litre of ${O_2}$ at STP?
[Please give your answer in the form of nearest integer]

Answer 1

From the given reaction, we can see that 2 moles of $KCl{O_3}$ are giving 3 moles of ${O_2}$. Volume of 1 mole of any substance at STP is 22.4 litres. So, by unitary method we can calculate the mass of $KCl{O_3}$ needed for $3 \times 22.4l$ of ${O_2}$.

Complete step by step solution:
Given reaction is:
$2KCl{O_3}\xrightarrow{{Mn{O_2}}}2KCl + 3{O_2}$
Now, we know that molecular mass is the sum of atomic masses of the elements present in a molecule. Molecular mass of a substance is calculated by multiplying the atomic masses of each element in the substance by the number of its atoms and adding them together.
Atomic mass of K = 39, atomic mass of Cl= 35.5, atomic mass of O = 16.
Now, molecular mass of $KCl{O_3}$ = ${\text{Atomic mass of K + Atomic mass of Cl + 3}} \times {\text{Atomic mass of O}}$
$\therefore$ Molecular mass of $KCl{O_3}$= $39 + 35.5 + 3 \times 16 = 122.5g$
Similarly molecular mass of KCl = $39 + 35.5 = 74.5g$
We also know that the volume of 1 mole of any substance at STP is 22.4 litres. Thus, the volume of 3 moles of ${O_2}$ is $3 \times 22.4l$.
It is clear from the given reaction that 2 moles of $KCl{O_3}$ are giving 3 moles of ${O_2}$. Therefore, $2 \times 122.5g$ of $KCl{O_3}$ is required to produce $3 \times 22.4l$ of ${O_2}$.
Or, $3 \times 22.4l$ of ${O_2}$ is produced by $KCl{O_3}$= $2 \times 122.5g$.
Hence, by unitary method,
6.72 litre of ${O_2}$ is produced by $KCl{O_3}$ = $\dfrac{{2 \times 122.5g}}{{3 \times 22.4l}} \times 6.72 = 24.5g$.

So, the mass of $KCl{O_3}$ required to produce 6.72 litre of ${O_2}$ at STP is 24.5 g. Hence, this is the required answer.

Note: At standard temperature and pressure (STP) condition, the volume occupied by one mole of any substance is called molar volume and its value is 22.4 litres. You can also be asked to calculate the number of molecules of any substance in the reaction. So you should know that 1 mole of any substance contains Avogadro's number i.e., $6.022 \times {10^{23}}$ of molecules.