Question

Find the no. of moles of ${O_2}$ produced from decomposition of 0.1 mole of $KCl{O_3}$?

Answer 1

Hint: Decomposition reaction is a reaction where only one reactant is decomposed to produce two or more products. So first write the decomposition reaction of $KCl{O_3}$ which is potassium chlorate. And find the no. of moles of oxygen produced, say x, originally from Potassium chlorate say y. y moles of $KCl{O_3}$ produces x moles of ${O_2}$, then find how many moles of ${O_2}$ produced from 0.1 mole of $KCl{O_3}$.

Complete step-by-step solution:We are given to find the no. of moles of ${O_2}$ produced from decomposition of 0.1 mole of $KCl{O_3}$.
Potassium chlorate $KCl{O_3}$ is strongly heated to break it down into an ionic potassium compound called potassium chloride releasing oxygen gas.
The reaction of the above process is $2KCl{O_3}\mathop \to \limits^{heat} 2KCl + 3{O_2} \uparrow $
As we can see 2 moles of potassium chlorate gives 2 moles of potassium chloride and 3 moles of oxygen gas.
As when 0.1 moles of potassium chlorate is heated it gives m moles of potassium chloride and n moles of oxygen gas.
The value of m will be equal to 0.1 as the no. of moles of potassium chlorate and potassium chloride is equal.
The value of n will be
$\begin{array}{*{20}{c}}
  {2moles\ of \ KCl{O_3}}& \to &{3moles\ of\ {O_2}} \\
  {0.1moles\ of\ KCl{O_3}}& \to &n
\end{array}$
$\Rightarrow n = \dfrac{{3 \times 0.1}}{2} = \dfrac{{0.3}}{2} = 0.15$

Therefore, 0.1 mole of $KCl{O_3}$ gives 0.15 moles of ${O_2}$ gas.

Note: Here initially for the reaction, we have taken 2 moles of $KCl{O_3}$ to get an integer no. of moles of ${O_2}$. Suppose if we take 1 mole of $KCl{O_3}$, then we get $\dfrac{3}{2}$ moles of ${O_2}$. But fraction no. of moles might confuse us a little, so always balance the reaction in such a way that the no. of moles of all the reactants and products must be integers.