Đáp án:
\(\begin{array}{l}
\% {m_{C{H_4}}} = 22,22\% \\
\% {m_{{C_2}{H_4}}} = 77,78\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = \dfrac{{50}}{{100}} = 0,5mol\\
{n_X} = \dfrac{{6,72}}{{22,4}} = 0,3\,mol\\
\left\{ \begin{array}{l}
a + b = 0,3\\
a + 2b = 0,5
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,2\\
\% {m_{C{H_4}}} = \dfrac{{0,1 \times 16}}{{0,1 \times 16 + 0,2 \times 28}} \times 100\% = 22,22\% \\
\% {m_{{C_2}{H_4}}} = 100 - 22,22 = 77,78\%
\end{array}\)
