Cho hơi nước đi qua than nóng đỏ, thu được 15,68 lít hỗn hợp khí X...

${{n}_{X}}=\dfrac{15,68}{22,4}=0,7 mol; {{n}_{NO}}=\dfrac{8,96}{22,4}=0,4 mol$
$H_{2}^{z}O\xrightarrow[t{}^\circ ]{+C}\left\{ \begin{aligned}
& \overset{x}{\mathop{CO}} , {{\overset{y}{\mathop{CO}} }_{2}} \\
& {{\overset{z}{\mathop{H}} }_{2}} \\
\end{aligned} \right.\xrightarrow[t{}^\circ ]{+CuO}\left\{ \begin{aligned}
& \left\{ \begin{aligned}
& Cu, \\
& CuO d\ddot{o} \\
\end{aligned} \right.\xrightarrow[{}]{+HN{{O}_{3}} d\ddot{o}}\left\{ \begin{aligned}
& Cu{{(N{{O}_{3}})}_{2}} \\
& NO \\
& {{H}_{2}}O \\
\end{aligned} \right. \\
& C{{O}_{2}},{{H}_{2}}O \\
\end{aligned} \right.$
$\to \left\{ \begin{aligned}
& {{n}_{X}}=x+y+z=0,7 \\
& \xrightarrow{BTE cho th\acute{i} nghie\ddot{a}m 1}2x+4y=2z \\
& \xrightarrow{BTE (\tilde{n}a\grave{a}u\to cuo\acute{a}i)}4(x+y)=3.0,4 \\
\end{aligned} \right.\to \left\{ \begin{aligned}
& x=0,2 \\
& y=0,1 \\
& z=0,4 \\
\end{aligned} \right.$
$\to \%{{V}_{CO}}=\dfrac{0,2}{0,7}.100=28,57\%$