\(n_{CH_4}=\dfrac{2}{16}=0,125\left(mol\right)\)
\(PTHH:CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(\left(mol\right)\) \(0,125\) \(0,25\)
Đặt \(\left\{{}\begin{matrix}n_{KMnO_4}=a\left(mol\right)\\n_{KClO_3}=b\left(mol\right)\end{matrix}\right.\)
\(\%m_K=26,68\left(\%\right)\Leftrightarrow\dfrac{39\left(a+b\right)}{158a+122,5b}=\dfrac{26,68}{100}\)
Lại có: \(0,5a+1,5b=0,25\) ( Cái này viết PTHH ra mới thấy)
\(\Rightarrow\left\{{}\begin{matrix}a=0,2\\b=0,1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\%m_{KMnO_4}=72\left(\%\right)\\\%m_{KClO_3}=28\left(\%\right)\end{matrix}\right.\)
$CH_4 + 2O_2 \xrightarrow{t^o} CO_2 + 2H_2O$
$n_{O_2} = 2n_{CH_4} = 2.\dfrac{2}{16} = 0,25(mol)$
Gọi $n_{KMnO_4} = a(mol) ; n_{KClO_3} = b(mol)$
Ta có :
$\dfrac{39(a + b)}{158a + 122,5b} = \dfrac{26,68}{100}(1)$
$2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2$
$2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2$
$n_{O_2} = 0,5a + 1,5b = 0,25(2)$
Từ (1)(2) suy ra a = 0,19 ; b = 0,1$
Ta có :
$\%m_{KMnO_4} = \dfrac{0,19.158}{0,19.158 + 0,1.122,5}.100\% = 71\%$
$\%m_{KClO_3} = 100\% - 71\% = 29\%$
