\(n_{O_2}=\dfrac{89,6}{22,4}=4mol\)
2H2+O2\(\overset{t^0}{\rightarrow}\)2H2O
2CO+O2\(\overset{t^0}{\rightarrow}\)2CO2
\(n_{H_2}=x;n_{CO}=y\)
Ta có hệ: \(\left\{{}\begin{matrix}2x+28y=68\\\dfrac{x}{2}+\dfrac{y}{2}=4\end{matrix}\right.\)
Giải ra x=6 và y=2
\(m_{H_2}=2.6=12gam\)
\(m_{CO}=2.28=56gam\)
\(V_{H_2}=6.22,4=134,4l\)
\(V_{CO}=2.22,4=44,8l\)