Đáp án:
\(\begin{array}{l}
\% {m_{C{H_4}}} = 63,16\% \\
\% {m_{{C_2}{H_4}}} = 36,84\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = \dfrac{{11,2}}{{22,4}} = 0,5\,mol\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
16a + 28b = 7,6\\
a + 2b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,3;b = 0,1\\
\% {m_{C{H_4}}} = \dfrac{{0,3 \times 16}}{{7,6}} \times 100\% = 63,16\% \\
\% {m_{{C_2}{H_4}}} = 100 - 63,16 = 36,84\%
\end{array}\)