Question

A sample of $KCl{O_3}$ on decomposition yielded $448ml$ of oxygen at NTP. Calculate
A.weight of the oxygen product
B.weight of $KCl{O_3}$ originally taken
C.weight of $KCl$ produced.
$2KCl{O_3}\xrightarrow{\Delta }2KCl + 3{O_2}$
$[K = 39,C = 25.5,O = 16]$

Answer 1

Volume at NTP is $22400ml$. And the formula to calculate the number of moles at NTP if volume is given is: number of moles is equal to the volume of a given compound in ml divided by volume at NTP i.e. $22400ml$.

Complete step by step answer:
In this question, we are given the decomposition of $KCl{O_3}$ at NTP. The decomposition of $KCl{O_3}$ is as: $2KCl{O_3}\xrightarrow{\Delta }2KCl + 3{O_2}$. This means if we are taking two moles of $KCl{O_3}$then on heating two moles of $KCl$and three moles of ${O_2}$will produce after the reaction.
Now, in the question we are given with volume of oxygen produced after the reaction which is $448ml$.
So, the number of moles of oxygen produced after the reaction will be $= \dfrac{{{\text{volume(in ml)}}}}{{{\text{volume at NTP}}}} = \dfrac{{448}}{{22400}} = 0.02$ ${\text{moles}}$.
We know that the mass of one mole of oxygen is $32grams$.
So the mass of $0.02moles$ of oxygen produced after the reaction will be $0.64grams$.
Now, if we take the moles of $KCl{O_3}$ as two moles, the mass of $KCl{O_3}$ will be moles multiplied by the molecular mass.
Molar mass of $KCl{O_3}$ $= 39 + 35.5 + 16 \times 3 = 122.5$
Hence the mass of $KCl{O_3}$will be $2 \times 122.5 = 245grams$.
We know that this weight of $KCl{O_3}$ will produce three moles of oxygen after the reaction.
So the weight of $KCl{O_3}$ required to produce $0.02$ moles of oxygen after the reaction will be $= \dfrac{{245 \times 0.02}}{3} = 1.63grams$.
Now we know that the mass before and after the reaction remains the same. According to the law of conservation of mass, the mass after and before remains the same i.e. mass neither be destroyed nor created. They only transfer from one compound to another.
So applying the law of conservation of mass ,
the mass of $KCl$ produced after the reaction will be as the mass of $KCl{O_3} -$mass of ${O_2}$ produced.
Mass of $KCl =$ $1.63 - 0.64 = 0.99grams$.
Weight of oxygen produced is $0.64grams$
Weight of $KCl{O_3}$ initially taken is as $1.63grams$
Weight of $KCl$ produced is $0.99grams$..

Note:
We know that by the law of conservation of mass that mass can neither be created nor be destroyed but they can change from one to another. So the total mass in the reaction remains the same before and after the reaction.