\(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ PTHH:2KClO_3\underrightarrow{t^o}2KCl+3O_2\\ \left(mol\right).....0,1..........0,1\leftarrow0,15\\ m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)
Ta có: \(n_{O_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)
PTHH: \(2KClO_3\xrightarrow[t^o]{V_2O_5}2KCl+3O_2\)
Theo PT: \(n_{KClO_3}=\dfrac{2}{3}.n_{O_2}=\dfrac{2}{3}.0,15=0,1\left(mol\right)\)
=> \(m_{KClO_3}=0,1.122,5=12,25\left(g\right)\)