Question

Oxygen is prepared by catalytic; decomposition of potassium chlorate $\left( {{\text{KCl}}{{\text{O}}_3}} \right)$. Decomposition, of potassium chlorate gives potassium chloride $\left( {{\text{KCl}}} \right)$ and oxygen $\left( {{{\text{O}}_2}} \right)$. How many moles and how many grams of ${\text{KCl}}{{\text{O}}_3}$ are required to produce $2.4$ mole ${{\text{O}}_2}$?

Answer 1

To determine the number of moles and grams of any reactant or product balanced equation is required. After writing the balanced equation, by comparing the number of moles of reactant and product, the amount of potassium chlorate can be determined. We can determine the number of gram weights by using the mole formula.

Formula used: ${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$

Complete step by step answer:
Decomposition, of potassium chlorate gives potassium chloride$\left( {{\text{KCl}}} \right)$ and oxygen $\left( {{{\text{O}}_2}} \right)$.
The reaction of decomposition of potassium chlorate is as follows:
${\text{2}}\,{\text{KCl}}{{\text{O}}_3}{\text{(s)}} \to 2{\text{KCl}}\,{\text{(s)}} + \,3\,{{\text{O}}_2}{\text{(g)}}\,\,$
According to the balanced reaction, the decomposition of two moles of potassium chlorate gives two moles of potassium chloride, and three moles of oxygen.
Compare the mole ratio to determine the mole of potassium chlorate required to reacts with $2.4$ mole oxygen as follows:
$3\,{\text{mol}}\,{{\text{O}}_2}\, = \,{\text{2}}\,\,{\text{mol}}\,\,{\text{KCl}}{{\text{O}}_3}$
$\Rightarrow 1\,{\text{mol}}\,{{\text{O}}_2}\, = \,\dfrac{2}{3}\,\,{\text{mol}}\,\,{\text{KCl}}{{\text{O}}_3}$
$\Rightarrow 2.4\,{\text{mol}}\,{{\text{O}}_2} = \dfrac{2}{3} \times 2.4 \, = \,1.6\,\,{\text{mol}}\,\,{\text{KCl}}{{\text{O}}_3}$
So, $1.6\,\,{\text{mol}}\,$ of ${\text{KCl}}{{\text{O}}_3}$ will give $2.4$ mole ${{\text{O}}_2}$.
We will the mole formula to determine the gram of ${\text{KCl}}{{\text{O}}_3}$, required to produce $2.4$ mole ${{\text{O}}_2}$ as follows:
${\text{Mole}}\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{{\text{Molar}}\,{\text{mass}}}}$
Molar mass of the potassium chlorate is $122.5\,{\text{g/mol}}$.
Substitute $122.5\,{\text{g/mol}}$for molar mass and $1.6\,\,{\text{mol}}\,$ for moles of potassium chlorate.
$\Rightarrow {\text{1}}{\text{.6}}\,{\text{mol}}\,\,{\text{ = }}\,\dfrac{{{\text{Mass}}}}{{122.5\,{\text{g/mol}}}}$
$\Rightarrow {\text{Mass}}\,{\text{of KCl}}{{\text{O}}_3}\,{\text{ = }}\,{\text{1}}{\text{.6}}\,{\text{mol}}\,\, \times 122.5\,{\text{g/mol}}$
$\Rightarrow {\text{Mass}}\,{\text{of KCl}}{{\text{O}}_3}\,{\text{ = }}\,{\text{1}}96\,{\text{g}}$
So, ${\text{1}}96\,{\text{g}}$ grams of ${\text{KCl}}{{\text{O}}_3}$ will give $2.4$ mole ${{\text{O}}_2}$.

**Therefore, $1.6\,\,{\text{mol}}\,$ and ${\text{1}}96\,{\text{g}}$ of ${\text{KCl}}{{\text{O}}_3}$ are required to produce $2.4$ moles ${{\text{O}}_2}$.

Note: **
The laboratory method consists of heating of potassium chlorate and manganese dioxide, evolves oxygen at $650\,{\text{K}}$ high temperature. At $650\,{\text{K}}$ temperature, potassium chlorate melts and is converted into potassium perchlorate which decomposes at higher temperature to evolve oxygen. Stoichiometry measurements are used to determine the amount of reactant or product from the given amounts. Stoichiometry measurements give quantitative relations among the amounts of various species of a reaction. To determine the stoichiometry relations a balanced equation is necessary.