Question

Calculate the mass of $KCl{ O }_{ 3 }$ required to liberate 6.72 $ { dm }^{ 3 }$ of oxygen at STP.
[Molar mass of potassium chlorate is 122.5g/mol]

Answer 1

Hint: Since there is no reaction given, we can assume that the question is talking about the decomposition of $KCl{ O }_{ 3 }$ that liberates ${ O }_{ 2 }$.

Complete step by step answer:
We will assume that the complete thermal decomposition of $KCl{ O }_{ 3 }$ is taking place. This reaction takes place at a temperature of 150-300 $^{ o }{ C }$ in the presence of $Mn{ O }_{ 2 }$ as a catalyst. The complete reaction for the decomposition of potassium chlorate is:
${ 2KClO }_{ 3 }(s)\xrightarrow [ 150-300^{ o }{ C } ]{ Mn{ O }_{ 2 } } 2KCl(s)+{ 3O }_{ 2 }(g)$

Hence it's clearly visible that 2 moles of $KCl{ O }_{ 3 }$ is required to produce 3 moles of ${ O }_{ 2 }$.
Now the molar volume at STP for an ideal gas is 22.4 L/mol. Since the question is asking the amount of potassium chlorate required to liberate oxygen gas at STP conditions, we are going to assume that the oxygen is behaving ideally at the STP conditions. Since one mole of ideal gas occupies 22.4 L volume, therefore for 6.72 L of ideal gas at STP conditions we will have 0.3 mole of that gas. The calculations are shown below:
1 mole of an ideal gas occupies 22.4 L volume at STP conditions therefore,
6.72 L of an ideal gas at STP = $\cfrac { 1\quad mole\quad of\quad gas\quad at\quad STP }{ 22.4L } \times 6.72L= 0.3mole$
Since 3 mole of ${ O }_{ 2 }$ is produced from 2 mole of $KCl{ O }_{ 3 }$, therefore 0.3 mole of ${ O }_{ 2 }$ will be produced from 0.2 mole of $KCl{ O }_{ 3 }$. The calculations are shown below:
3 moles of ${ O }_{ 2 }$ is produced from = 2 moles of $KCl{ O }_{ 3 }$
0.3 moles of ${ O }_{ 2 }$ is produced from = $\cfrac { 2\quad mole\quad of\quad KCl{ O }_{ 3 } }{ 3\quad mole\quad of\quad { O }_{ 2 } } \times 0.3\quad mole\quad of\quad { O }_{ 2 }$
Therefore, 0.3 moles of ${ O }_{ 2 }$ is produced from 0.2 moles of $KCl{ O }_{ 3 }$

Now, Mass of a substance = $Molar\quad mass\times number\quad of\quad moles$
The Molar mass of $KCl{ O }_{ 3 }$ is given as 122.5 g/mol, therefore the mass of $KCl{ O }_{ 3 }$ required will be 24.5g. The calculation is shown below:
Mass of $KCl{ O }_{ 3 }$ = $122.5gmo{ l }^{ -1 }\times 0.2\quad mole=24.5g$
Hence the mass required will be 24.5 g.

Note: Do not get confused between STP and NTP conditions. In STP (Standard temperature and pressure) condition, the temperature is 298.15 K and the pressure is 1 bar. Whereas in NTP (Normal temperature and pressure) condition, the temperature is 293.15 K and the pressure is 101.325 kPa.