\(2KClO_3\underrightarrow{t^0}2KCl+3O_2\uparrow\)
2 mol 3 mol
\(\dfrac{a}{122,5}\) mol \(\dfrac{3a}{2.122,5}\)mol
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+O_2\uparrow+MnO_2\)
2 mol 1 mol
\(\dfrac{b}{158}\)mol \(\dfrac{b}{2.158}\)mol
Muốn được cùng một lượng oxi: \(\dfrac{3a}{2.122,5}=\dfrac{b}{2.158}\)
Rút ra tỷ lệ: \(\dfrac{a}{b}=\dfrac{245}{958}=\dfrac{7}{27,0875}\)
\(n_{KClO_3}=\dfrac{a}{122,5}\left(mol\right)\)
\(n_{KMnO_4}=\dfrac{b}{158}\left(mol\right)\)
PTHH: \(2KClO_3\underrightarrow{t^0}2KCl+3O_2\uparrow\left(1\right)\)
\(2KMnO_4\underrightarrow{t^0}K_2MnO_4+MnO_2+O_2\uparrow\left(2\right)\)
Theo PT (1): \(n_{O_2}=\dfrac{3}{2}.n_{KClO_3}=\dfrac{3}{2}.\dfrac{a}{122,5}=\dfrac{3a}{245}\left(mol\right)\)
Theo PT (2): \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}.\dfrac{b}{158}=\dfrac{b}{316}\left(mol\right)\)
\(\Rightarrow\left\{{}\begin{matrix}m_{O_2\left(1\right)}=\dfrac{3a}{245}.32=\dfrac{96a}{245}\left(g\right)\\m_{O_2\left(2\right)}=\dfrac{b}{316}.32=\dfrac{8b}{79}\left(g\right)\end{matrix}\right.\)
Theo đề bài khối lượng thu được ở 2 phản ứng bằng nhau
\(\Rightarrow\dfrac{96a}{245}=\dfrac{8b}{79}\Rightarrow96a.79=8b.245\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{245}{948}\)