Giả sử lấy cùng 1 khối lượng là a gam.
\(n_{KClO_3}=\dfrac{m_{KClO_3}}{M_{KClO_3}}=\dfrac{a}{122,5}\left(mol\right)\)
\(n_{KMnO_4}=\dfrac{m_{KMnO_4}}{M_{KMnO_4}}=\dfrac{a}{158}\left(mol\right)\)
PT (1):
\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\)
2mol............2mol........3mol
\(\dfrac{a}{158}\)mol.......................\(\dfrac{3a}{245}\)
Theo PT (1) ta có: \(n_{O_2}=\dfrac{3.a}{122,5.2}=\dfrac{3a}{245}\)
\(\Rightarrow V_{O_2\left(1\right)}=\dfrac{3a}{245}.22,4=\dfrac{48a}{175}\left(l\right)\)
PT (2):
\(2KMnO_4\underrightarrow{t^o}K_2MnO_4+MnO_2+O_2\uparrow\)
2mol.............1mol...............1mol.........1mol
\(\dfrac{a}{158}\)mol..............................................\(\dfrac{a}{316}\)mol
Theo PT (2) ta có: \(n_{O_2}=\dfrac{a}{158.2}=\dfrac{a}{316}\)
\(\Rightarrow V_{O_2\left(2\right)}=\dfrac{a}{316}.22,4=\dfrac{28a}{395}\left(l\right)\)
\(V_{O_2\left(1\right)}=\dfrac{48a}{175}>V_{O_2\left(2\right)}=\dfrac{28a}{395}\Rightarrow KClO_3ChoTheTichO2Hon\)
