Đáp án:

\(\begin{array}{l}
a)\\
{m_{CO}} = 2,8g\\
{m_{{C_3}{H_4}}} = 4g\\
b)\\
\% {V_{CO}} = 50\% \\
\% {V_{{C_3}{H_4}}} = 50\% 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\\
2CO + {O_2} \xrightarrow{t^0} 2C{O_2}\\
{C_3}{H_4} + 4{O_2} \xrightarrow{t^0} 3C{O_2} + 2{H_2}O\\
hh:CO(a\,mol),{C_3}{H_4}(b\,mol)\\
{n_{hh}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
{n_{C{O_2}}} = \dfrac{{17,6}}{{44}} = 0,4\,mol\\
\left\{ \begin{array}{l}
a + b = 0,2\\
a + 3b = 0,4
\end{array} \right.\\
 \Rightarrow a = b = 0,1\\
{m_{CO}} = 0,1 \times 28 = 2,8g\\
{m_{{C_3}{H_4}}} = 0,1 \times 40 = 4g\\
b)\\
\% {V_{CO}} = \dfrac{{0,1 \times 22,4}}{{4,48}} \times 100\%  = 50\% \\
\% {V_{{C_3}{H_4}}} = 100 - 50 = 50\% 
\end{array}\)