Giải thích các bước giải:
$n_X=$ $\dfrac{11,2}{22,4}=0,5(mol)$
$Do$ $d_{X/N_2}=2→ \overline{M_X}=2*28=56(g/mol)$
$CO_2:44$ $8$
$56$
$SO_2: 64$ $12$
$→$ $\dfrac{n_{CO_2}}{n_{SO_2}}=$ $\dfrac{8}{12}=$ $\dfrac{2}{3}$
$\rightarrow \left\{\begin{matrix} n_{CO_2}=0,2(mol) & \\ n_{SO_2}=0,3(mol) & \end{matrix}\right.$
$\rightarrow \left\{\begin{matrix} \%V_{CO_2}=\dfrac{0,2}{0,5}=40\% & \\ \%V_{SO_2}=100\% -40\%=60\% & \end{matrix}\right.$
$\left\{\begin{matrix}
m_{CO_2}=0,2*44=8,8(g) & \\
m_{SO_2}=0,3*64=19,2(g) &
\end{matrix}\right.
\rightarrow \left\{\begin{matrix}
\%m_{CO_2}=\dfrac{8,8}{8,8+19,2}*100\%= 31,43\%& \\
\%m_{SO_2}=100\%-31,43\%=68,57\% &
\end{matrix}\right.$
