Em tham khảo nha:
\(\begin{array}{l}
{n_{hh}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
\text{ Bình 1 }hh:C{O_2}(a\,mol),{H_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,5\\
44a + 2b = 7,5
\end{array} \right.\\
\Rightarrow a = \dfrac{{13}}{{84}};b = \dfrac{{29}}{{84}}\\
\% {V_{C{O_2}}} = \dfrac{{\frac{{13}}{{84}}}}{{0,5}} \times 100\% = 30,95\% \\
\% {V_{{H_2}}} = 100 - 30,95 = 69,05\% \\
\text{ Bình 2} hh:CO(x\,mol),{H_2}(y\,mol)\\
\left\{ \begin{array}{l}
x + y = 0,5\\
28x + 2y = 7,5
\end{array} \right.\\
\Rightarrow x = y = 0,25\\
\% {V_{CO}} = \% {V_{{H_2}}} = \dfrac{{0,25}}{{0,5}} \times 100\% = 50\%
\end{array}\)