Đáp án:
\(\begin{array}{l}
\% C{H_4} = 48,8\% \\
\% {C_2}{H_4} = 51,2\% \\
{V_{{O_2}}} = 42,56l
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
{n_{C{O_2}}} = \dfrac{{48,4}}{{44}} = 1,1mol\\
{n_{hh}} = \dfrac{{17,92}}{{22,4}} = 0,8mol\\
hh:C{H_4}(a\,mol);{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,8\\
a + 2b = 1,1
\end{array} \right.\\
\Rightarrow a = 0,5;b = 0,3\\
{m_{C{H_4}}} = 0,5 \times 16 = 8g\\
{m_{{C_2}{H_4}}} = 0,3 \times 28 = 8,4g\\
\% C{H_4} = \dfrac{8}{{8 + 8,4}} \times 100\% = 48,8\% \\
\% {C_2}{H_4} = 100 - 48,8 = 51,2\% \\
{n_{{O_2}}} = 2{n_{C{H_4}}} + 3{n_{{C_2}{H_4}}} = 1,9mol\\
{V_{{O_2}}} = 1,9 \times 22,4 = 42,56l
\end{array}\)
