$2KClO_{3} \underrightarrow{t^{o}} 2KCl + 3O_{2}\uparrow$
$n_{KClO_{3}}=\frac{18,375}{122,5}=0,15 (mol)$
$\text{Theo ptpư:}\text{ }n_{O_{2}}=\frac{3}{2}n_{KClO_{3}}=0,225 (mol)$
$\Rightarrow V_{O_{2}}=0,225.22,4=5,04 (l)$
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