Đáp án:
\(\begin{array}{l}
\% {V_{{C_2}{H_2}}} = 44,4\% \\
\% {V_{{C_2}{H_4}}} = 55,6\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
{n_{hh}} = \dfrac{{10,08}}{{22,4}} = 0,45mol\\
{n_{{H_2}O}} = \dfrac{{12,6}}{{18}} = 0,7mol\\
hh:{C_2}{H_2}(a\,mol),{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,45\\
a + 2b = 0,7
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,25\\
\% {V_{{C_2}{H_2}}} = \dfrac{{0,2 \times 22,4}}{{10,08}} \times 100\% = 44,4\% \\
\% {V_{{C_2}{H_4}}} = 100 - 44,4 = 55,6\%
\end{array}\)