\(m_{KClO_3}=32,67.\dfrac{100-25}{100}=24,5025\left(g\right)\)
=> \(n_{KClO_3}=\dfrac{24,5025}{122,5}=0,20\left(mol\right)\)
PTHH: \(2KClO_3-t^o->2KCl+3O_2\)
Theo PT ta có: \(n_{O_2}=\dfrac{0,20.3}{2}=0,3\left(mol\right)\)
=> \(V_{O_2\left(đktc\right)}=0,3.22,4=6,72\left(l\right)\)
Theo PT ta có: \(n_{KCl}=n_{KClO_3}=0,20\left(mol\right)\)
=> \(m_{KCl}=0,20.74,5=14,9\left(g\right)\)