Answer
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Hint: A balance chemical is that when the stoichiometric coefficient of a reactant is equal to the product To balance the equation we must make sure that both sides of the reaction have equal numbers of atoms. For this we must first try to balance the oxygen atoms and then the other atoms will be added later.
Complete step by step answer:
Potassium chlorate undergoes decomposition to form potassium chloride and oxygen gas. The reaction can be written as follows,
$KClO3\to KCl+O2$
We can see that the oxygen on the reactant side is more than on the product side. For this reason, we will add $2$to the reactant side that is, we will allow 2 molecules of potassium chlorate to undergo the reaction.
$2KClO3\to KCl+O2$
Now we have to balance the product side by adding $2$ in the place of the stoichiometric coefficient of the $KCl$. The equation is as follows:
$2KClO3\to 2KCl+O2$
Now we have to balance the oxygen atoms. We can see that on the reactant side, 6 oxygens are present. Therefore, we will add $3$ as a coefficient for oxygen.
$2KClO3\to 2KCl+3O2$ .Thus, we now have a balanced equation.
Note: We must make sure that while we are balancing the reaction, we cannot change the molecular formula of the compounds. Hence when we are adding a number in the place of the stoichiometric coefficient we are not changing the molecular formula of the compound.
We can also add fractions for a stoichiometric coefficient. It is commonly used in reactions where combustion takes place. For example:
$C+\dfrac{1}{2}O2\to CO$
Potassium chlorate is a crystalline solid which is white in colour. It is highly inflammable and can be ignited with friction itself. It reacts very strongly with sulphuric acid and can cause an explosion. It has an
Complete step by step answer:
Potassium chlorate undergoes decomposition to form potassium chloride and oxygen gas. The reaction can be written as follows,
$KClO3\to KCl+O2$
We can see that the oxygen on the reactant side is more than on the product side. For this reason, we will add $2$to the reactant side that is, we will allow 2 molecules of potassium chlorate to undergo the reaction.
$2KClO3\to KCl+O2$
Now we have to balance the product side by adding $2$ in the place of the stoichiometric coefficient of the $KCl$. The equation is as follows:
$2KClO3\to 2KCl+O2$
Now we have to balance the oxygen atoms. We can see that on the reactant side, 6 oxygens are present. Therefore, we will add $3$ as a coefficient for oxygen.
$2KClO3\to 2KCl+3O2$ .Thus, we now have a balanced equation.
Note: We must make sure that while we are balancing the reaction, we cannot change the molecular formula of the compounds. Hence when we are adding a number in the place of the stoichiometric coefficient we are not changing the molecular formula of the compound.
We can also add fractions for a stoichiometric coefficient. It is commonly used in reactions where combustion takes place. For example:
$C+\dfrac{1}{2}O2\to CO$
Potassium chlorate is a crystalline solid which is white in colour. It is highly inflammable and can be ignited with friction itself. It reacts very strongly with sulphuric acid and can cause an explosion. It has an
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