https://hoc24.vn/hoi-dap/question/43901.html
2KClO3 → 2KCl + 3O2 (1)
2KMnO4 → K2MnO4 + MnO2 + O2 (2)
a)Theo PT1:
nKClO3= \(\frac{a}{122,5}\)(mol)
nKCl= nKClO3= \(\frac{a}{122,5}\)(mol)
Theo PT2:
nKMnO4= \(\frac{b}{158}\)(mol)
nK2MnO4= 2nKMnO4= 2.\(\frac{b}{158}\)= \(\frac{b}{79}\)(mol)
nMnO2= 2nKMnO4= 2.\(\frac{b}{158}\)= \(\frac{b}{79}\)(mol)
Theo đề bài, ta có phương trình:
\(\frac{a}{122,5}\).74,5 = \(\frac{b}{79}\). 197 + \(\frac{b}{79}\).87
Tỉ lệ: \(\frac{a}{b}\)= \(\frac{122,5.\left(197+87\right)}{2.158.74,5}\)
➩\(\frac{a}{b}\)≃1,78
b) Theo đề bài ta có:
VO2(1) : VO2(2)= \(\frac{3a}{2}\).22,4 : \(\frac{b}{2}\).22,4=3.\(\frac{a}{b}\)= 4,43