\(n_X=\frac{1,12}{22,4}=0,5\left(mol\right)\)
\(d_{X/O2}=0,325\Rightarrow\overline{M_X}=0,325.32=10,4\left(\frac{g}{mol}\right)\)
Áp dụng hệ quả quy tắc đường chéo:
\(\frac{n_{H2}}{n_{CH4}}=\frac{16-10,4}{10,4-2}=\frac{2}{3}\)
\(\Rightarrow n_{H2}=0,2\left(mol\right)\)
\(n_{CH4}=0,3\left(mol\right)\)
Phản ứng xảy ra:
\(2H_2+O_2\rightarrow2H_2O\)
\(CH_4+2O_2\rightarrow CO_2+2H_2O\)
\(n_{O2}=\frac{17,92}{22,4}=0,8\left(mol\right)\)
Theo PTHH
\(\Rightarrow n_{O2\left(pư\right)}=\frac{1}{2}.n_{H2}+2n_{CH4}=\frac{1}{2}.0,2+2.0,3=0,7\left(mol\right)\)
\(\Rightarrow n_{O2\left(dư\right)}=0,8-0,7=0,1\left(mol\right)\)
\(n_{CO2}=n_{CH4}=0,3\left(mol\right)\)
Vậy Y gồm \(\left\{{}\begin{matrix}0,1\left(mol\right)O_2\left(dư\right)\\0,3\left(mol\right)CO_2\end{matrix}\right.\)
\(\Rightarrow\%V_{O2\left(dư\right)}=\frac{0,1}{0,1+0,3}.100\%=25\%\)
\(\Rightarrow\%V_{CO2\left(Y\right)}=100\%-25\%=75\%\)
