Question

Calculate the total number of moles of atoms of each element present in $ 122.5{\text{g}} $ of $ KCl{O_3} $ .

Answer 1

Hint: We need to calculate the atomic mass of $ KCl{O_3} $ . Further we need to apply the mole formula. Then we shall calculate the moles of the compound given and thus, the moles of each element.

Formula used:
 $ n = \dfrac{{{\text{weight of substance}}}}{{{\text{molar mass of substance}}}} $
Here, $ n $ is the number of moles.

Complete step by step answer
We already know the molar masses of $ KCl{O_3} $ which is $ 122.5{\text{g}} $
We will find out the number of moles,
 $ {\text{Moles of KCl}}{{\text{O}}_{\text{3}}} = \dfrac{{122.5}}{{122.5}} = 1{\text{ moles}} $
Hence it contains 1 mole of K atom, 1 mole of Cl atom and 3 moles of O atom.
So, total no of moles of each element are calculated.

Note
This type of question requires a lot of calculation which should be done very carefully especially when taking the no of moles.
Some facts about potassium chlorate is a strong oxidizing agent that has a wide variety of uses. It is or has been a component of explosives, fireworks, safety matches, and disinfectants. As a high school student, we all have used it to generate oxygen in the lab. When it is heated strongly, it breaks down, releasing oxygen gas and leaving behind a thermally stable solid residue of an ionic potassium compound. There are at least three plausible reactions one can write for the process, but only one occurs to any significant extent.
Also, here the concept of limiting agent comes. It is the reactant in a chemical reaction that limits the amount of product that can be formed. The reaction will stop when all of the limiting reactant is consumed whereas excess reactant is the reactant in a chemical reaction that remains when a reaction stops when the limiting reactant is completely consumed.