Question

A given initial mass of $KCl{O_3}$ on $50\% $ decomposition produces $67.2$ litre oxygen at $0^\circ C$ and $1atm$, the other product of decomposition is $KCl$, the initial mass of $KCl{O_3}$( in $gm$) taken is
A. $245$
B. $122.5$
C. $490$
D. None of these

Answer 1

Hint: we know that in the decomposition reaction of $KCl{O_3}$, $KCl$ and oxygen gas are formed. In the question the value of volume of oxygen gas is given at $0^\circ C$ and $1atm$ which is also known as the standard temperature and pressure condition. We will use the mole method to solve the question.

Complete step by step solution:
We know that the decomposition reaction of $KCl{O_3}$ can be written as $KCl{O_3} \to KCl + {O_2}$. We will have to balance the decomposition chemical reaction , so after balancing we will get $2KCl{O_3} \to 2KCl + 3{O_2}$. The condition that is given in the question is that the temperature is $0^\circ C$ and the pressure is $1atm$. This condition is known as the $STP$ condition. In $STP$ condition one mole of a gas has volume of $22.4L$. In the question we have been given that $67.2$ litre oxygen is being produced. That means $ \Rightarrow \dfrac{{67.2}}{{22.4}} = 3$ moles of oxygen is produced. As in the chemical equation $2KCl{O_3} \to 2KCl + 3{O_2}$ we can see that two moles of $KCl{O_3}$ produces three moles of oxygen gas. If the reaction was $100\% $decomposed then $ \Rightarrow 2(30 + 35.5 + 48) = 245g$ of $KCl{O_3}$ would have been required because the mass of two moles of $KCl{O_3}$ is $245g$. But as we have been given in the question that the reaction is $50\% $ decomposed so we will require double the mass of the $KCl{O_3}$ to get $67.2$ litre oxygen, that is $(245 \times 2 = 490g)$. So from the above explanation and calculation it is clear to us that

The correct answer of the given question is option: C.

Additional information:
Always remember that when the temperature is $0^\circ C$ and pressure is $1atm$ it is called as $STP$ condition. In $STP$ condition the volume of one mole of a gas is always $22.4L$. The full form of $STP$is standard temperature and pressure.

Note: Always remember that the decomposition reaction of $KCl{O_3}$ is given by $2KCl{O_3} \to 2KCl + 3{O_2}$. The molecular mass of $KCl{O_3}$ is $ \Rightarrow (30 + 35.5 + 48) = 122.5g$. Always remember that the volume occupied by any gas in $STP$ condition is always $22.4L$.