Em tham khảo nha :

\(\begin{array}{l}
a)\\
hh:C{O_2}(a\,mol),CO(b\,mol)\\
{M_A} = 2,5{M_{C{H_4}}} = 40dvC\\
 \Rightarrow \dfrac{{44a + 28b}}{{a + b}} = 40\\
 \Rightarrow \dfrac{a}{b} = \dfrac{{40 - 28}}{{44 - 40}} = 3\\
\% C{O_2} = \dfrac{{3 \times 44}}{{3 \times 44 + 28}} \times 100\%  = 82,5\% \\
\% CO = 100 - 82,5 = 17,5\% \\
\% {V_{C{O_2}}} = \dfrac{{3 \times 22,4}}{{(3 + 1) \times 22,4}} \times 100\%  = 75\% \\
\% {V_{CO}} = 100 - 75 = 25\% \\
b)\\
{m_{CO}} = \dfrac{{16 \times 17,5}}{{100}} = 2,8g\\
2CO + {O_2} \to 2C{O_2}\\
{n_{CO}} = \dfrac{{2,8}}{{28}} = 0,1mol\\
{n_{{O_2}}} = \dfrac{{{n_{CO}}}}{2} = 0,05mol\\
{V_{{O_2}}} = \dfrac{{nRT}}{p} = \dfrac{{0,05 \times 0,082 \times (27 + 273)}}{{1,1}} = 1,11818l
\end{array}\)