Answer:
A: 9 mol
Explanation:
The reaction is
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
The stoichiometric relation is given by
[tex]\text{Number of moles of }KClO_3\times \dfrac{\text{Moles of}\ O_2}{\text{Moles of}\ KClO_3}\\ =\text{6 mol of}\ KClO_3\times \dfrac{\text{3 mol of }O_2}{\text{2 mol of }KClO_3}\\ =9\ \text{mol of }O_2[/tex]
The moles of oxygen gas produced in the given reaction is [tex]9\ \text{mol of }O_2[/tex].
