Gọi \(\left\{{}\begin{matrix}n_{O2}:a\left(mol\right)\\n_{O3}:b\left(mol\right)\end{matrix}\right.\)
\(\Leftrightarrow\frac{32a+48b}{a+b}=17,62\)
\(\Leftrightarrow12,8b=3,2a\Leftrightarrow a=4b\)
Vì thể tích tỉ lệ với số mol
\(\%V_{O2}=\frac{a}{a+b}.100\%=\frac{4b}{4b+b}.100\%=80\%\)
\(\Rightarrow\%V_{O3}=100\%-80\%=20\%\)
5,04 l khí A chứa:
\(\left\{{}\begin{matrix}n_{O2}=\frac{5,04.80\%}{22,4}=0,18\left(mol\right)\\n_{O3}=\frac{5,04.20\%}{22,4}=0,045\left(mol\right)\end{matrix}\right.\)
\(2O_3\underrightarrow{^{t^o}}3O_2\)
0,045_0,0675_
\(\Rightarrow\Sigma n_{O2}=0,18+0,0675=0,2475\left(mol\right)\)
\(CH_4+2O_2\rightarrow CO_2+H_2O\)
0,12375__0,2475 ______
\(\Rightarrow V_{CH4}=0,12375.22,4=2,772\left(l\right)\)
