Question

The mole(s) of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ which on decomposition give just sufficient ${{\text{O}}_{\text{2}}}$ to convert 2 mole of aluminium to its oxide, is/are
a.) 1.5
b.) 2.5
c.) 1
d.) 1.33

Answer 1

Hint: At first we should be writing down the equation which is mentioned in the question. Then we can find out the number of moles of ${\text{KCl}}{{\text{O}}_{\text{3}}}$.

Complete answer:
In the question it is given that,
2 moles of aluminium is oxidized.
Which can be expressed as:
$\text{2Al + }\dfrac{\text{3}}{\text{2}}{{\text{O}}_{\text{2}}}\text{ }\xrightarrow{{}}\text{ A}{{\text{l}}_{\text{2}}}{{\text{O}}_{\text{3}}}$

Which means, 1.5 moles of ${{\text{O}}_{\text{2}}}$ is used in oxidation of 2 moles of Al.
Again,
$\text{KCl}{{\text{O}}_{\text{3}}}\text{ }\xrightarrow{{}}\text{ KCl + }\dfrac{\text{3}}{\text{2}}{{\text{O}}_{\text{2}}}$
Where,

1 mole of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ is produce 1.5 mole of ${{\text{O}}_{\text{2}}}$.
Therefore, 1 mole of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ is used.

Therefore, the correct answer is 1 mole of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ is used for the decomposition, so as to give sufficient ${{\text{O}}_{\text{2}}}$ to convert 2 mole of aluminium to its oxide.

So, the correct option is Option C.

Additional Information: Chemical decomposition in chemistry, is defined as the process or effect of simplifying a single chemical entity into two or more fragments.
Chemical decomposition is usually defined as the opposite of the process of chemical synthesis. By chemical synthesis we mean the artificial execution of useful chemical reactions to obtain one or several products.

Note: The reaction of the decomposition of ${\text{KCl}}{{\text{O}}_{\text{3}}}$ takes place at a temperature of ${{150}^{\text{o}}}\text{C-30}{{\text{0}}^{\text{o}}}\text{C}$. When ${\text{KCl}}{{\text{O}}_{\text{3}}}$ is strongly heated, it breaks down, releasing oxygen gas and leaving behind a thermally stable solid residue of an ionic potassium compound.
By thermally stable we mean, heat insensitive.
The catalyst of this reaction is manganese (IV) oxide.