a) Ta có: \(M_A=20.2=40\left(g/mol\right)\)
Gọi \(\left\{{}\begin{matrix}n_{O_2}=a\left(mol\right)\\n_{O_3}=b\left(mol\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m_{O_2}=32a\left(g\right)\\m_{O_3}=48b\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}n_A=n_{O_2}+n_{O_3}=a+b\left(mol\right)\\m_A=m_{O_2}+m_{O_3}=32a+48b\left(g\right)\end{matrix}\right.\)
=> \(M_A=\dfrac{m_A}{n_A}=\dfrac{32a+48b}{a+b}=40\left(g/mol\right)\)
=> 32a + 48b = 40a + 40b
=> 8b = 8a
=> a = b
Ta có %V cũng là %n
=> \(\left\{{}\begin{matrix}\%V_{O_2}=\%n_{O_2}=\dfrac{a}{a+b}.100\%=\dfrac{a}{a+a}.100\%=\dfrac{1}{2}.100\%=50\%\\\%V_{O_3}=100\%-50\%=50\%\end{matrix}\right.\)
b) 1 mol khí có \(\left\{{}\begin{matrix}n_{O_2}=\dfrac{1.50}{100}=0,5\left(mol\right)\\n_{O_3}=1-0,5=0,5\left(mol\right)\end{matrix}\right.\)
PTHH: \(CH_4+2O_2\xrightarrow[]{t^o}CO_2+2H_2O\left(1\right)\)
\(3CH_4+4O_3\xrightarrow[]{t^o}3CO_2+6H_2O\left(2\right)\)
Theo PTHH (1), (2): \(n_{CH_4}=\dfrac{3}{4}.n_{O_3}+\dfrac{1}{2}.n_{O_2}=\dfrac{3}{4}.0,5+\dfrac{1}{2}.0,5=0,625\left(mol\right)\)
=> \(V_{CH_4}=0,625.22,4=14\left(l\right)\)
