Question

If potassium chlorate is $80\%$ pure, then $48{\text{ g}}$ of oxygen would be produced from (atomic mass of $K{\text{ = 39}}$ ):
A. ${\text{153}}{\text{.12 g of KCl}}{{\text{O}}_3}$
B. ${\text{122}}{\text{.5 g of KCl}}{{\text{O}}_3}$
C. ${\text{245 g of KCl}}{{\text{O}}_3}$
D. ${\text{98 g of KCl}}{{\text{O}}_3}$

Answer 1

We will write a balanced decomposition reaction of potassium chlorate. With the help of the reaction we will find the amount of oxygen produced on decomposition of one mole of potassium chlorate. Then by using the purity of potassium chlorate we will find the amount of potassium chloride needed to produce $48{\text{ g}}$ of oxygen.

Complete answer: Potassium chlorate will undergo a decomposition reaction to form potassium chloride and oxygen gas as by-products. We will find the amount of potassium chlorate required to produce $48{\text{ g}}$ of oxygen gas during the reaction. We are given a mass of oxygen produced which is equal to $48{\text{ g}}$ and the purity of potassium chlorate is $80\%$. Firstly we will write the decomposition reaction of potassium chlorate as,
$KCl{O_3}{\text{ }}\xrightarrow{{}}{\text{ KCl + }}{{\text{O}}_2}$
We will balance the above reaction to get an exact amount of potassium chlorate. Therefore the balanced decomposition reaction will be,
$2KCl{O_3}{\text{ }}\xrightarrow{{}}{\text{ 2KCl + 3}}{{\text{O}}_2}$
Molar mass of $KCl{O_3}{\text{ = 39 + 37}}{\text{.5 + 3 }} \times {\text{ 16}}$
Molar mass of $KCl{O_3}{\text{ = 122}}{\text{.5 g}}$
From the above reaction we can see that two moles of $KCl{O_3}$ will produce three moles of oxygen gas. We can write as,
$2{\text{ }} \times {\text{ 122}}{\text{.5 g }}\xrightarrow{{produces}}{\text{ 3 }} \times {\text{ 32 g}}$
${\text{245 g }}\xrightarrow{{produces}}{\text{ 96 g}}$
Therefore one gram of oxygen will be produced by $\dfrac{{245}}{{96}}{\text{ g }}$ of $KCl{O_3}$ but we are given that $48{\text{ g}}$ of oxygen is produced while decomposition. Therefore the amount of $KCl{O_3}$ consumed for producing $48{\text{ g}}$ of oxygen will be equal to $\dfrac{{245}}{{96}}{\text{ }} \times {\text{ 48 g}}$ which is equal to ${\text{122}}{\text{.5 g }}$. Thus the amount of $KCl{O_3}$ consumed will be ${\text{122}}{\text{.5 g }}$. But the purity of $KCl{O_3}$ is $80\%$ , thus we will assume the mass of mixture be ${\text{x g }}$, then mass of mixture will be calculated as,
$80\%$ Of ${\text{x g = 122}}{\text{.5 g}}$
${\text{x g = }}\dfrac{{{\text{122}}{\text{.5}}}}{{80}}{\text{ }} \times {\text{ 100 g}}$
${\text{x g = 153}}{\text{.125 g}}$
Thus the amount of mixture will be $153.125{\text{ g}}$. Therefore the correct answer is ${\text{153}}{\text{.12 g of KCl}}{{\text{O}}_3}$.

Note:
The decomposition reaction must be balanced before solving the question. We have used a unitary method to find the amount of potassium chlorate consumed for producing $48{\text{ g}}$ of oxygen. Since the purity of the mixture is given, it is mandatory to calculate the exact amount of potassium chlorate because from the reaction we have calculated for impure potassium chlorate.