$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
a)
$n_{O_2} = \dfrac{64}{32} = 2(mol)$
Theo PTHH :
$n_{KClO_3} = \dfrac{2}{3}n_{O_2} = \dfrac{4}{3}(mol)$
$m_{KClO_3} = \dfrac{4}{3}.122,5 = 163,3(gam)$
b) $n_{O_2} = \dfrac{7,437}{24,79} = 0,3(mol)$
$\Rightarrow n_{KClO_3} = 0,2(mol)$
$\Rightarrow m_{KClO_3} = 0,2.122,5 = 24,5(gam)$