Gọi x, y lần lượt là số mol của KMnO4 và KClO3
Theo đề, ta có: \(158x+122,5y=273,4\) (*)
Ta có: \(n_{O_2}=\dfrac{49,28}{22,4}=2,2\left(mol\right)\)
PTHH:
\(2KMnO_4\overset{t^o}{--->}K_2MnO_4+MnO_2+O_2\uparrow\left(1\right)\)
\(2KClO_3\xrightarrow[MnO_2]{t^o}2KCl+3O_2\uparrow\left(2\right)\)
Theo PT(1): \(n_{O_2}=\dfrac{1}{2}.n_{KMnO_4}=\dfrac{1}{2}x\left(mol\right)\)
Theo PT(2): \(n_{KClO_3}=\dfrac{3}{2}.n_{KClO_3}=\dfrac{3}{2}y\left(mol\right)\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{3}{2}y=2,2\) (**)
Từ (*) và (**), ta có HPT:
\(\left\{{}\begin{matrix}158x+122,5y=273,4\\\dfrac{1}{2}x+\dfrac{3}{2}y=2,2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=1,2\end{matrix}\right.\)
\(\Rightarrow m_{KMnO_4}=0,8.158=126,4\left(g\right)\)
\(\Rightarrow\%_{m_{KMnO_4}}=\dfrac{126,4}{273,4}.100\%=46,23\%\)
\(\%_{m_{KClO_3}}=100\%-46,23\%=53,77\%\)
\(n_{O_2}=\dfrac{49,28}{22,4}=2,2mol\)
\(2KMnO_4\rightarrow K_2MnO_4+MnO_2+O_2\)
\(x\) \(\dfrac{x}{2}\)
\(2KClO_3\rightarrow2KCl+3O_2\)
\(y\) \(\dfrac{3}{2}y\)
Ta có: \(\left\{{}\begin{matrix}158x+122,5y=273,4\\\dfrac{1}{2}x+\dfrac{3}{2}y=2,2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0,8mol\\y=1,2mol\end{matrix}\right.\)
\(\%m_{KMnO_4}=\dfrac{158\cdot0,8}{273,4}\cdot100\%=46,23\%\)
\(\%m_{KClO_3}=100\%-46,23\%=53,77\%\)