Đáp án:
\(\begin{array}{l}
\% C{H_4} = 80\% \\
\% {C_2}{H_2} = 20\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
2{C_2}{H_2} + 5{O_2} \to 4C{O_2} + 2{H_2}O\\
b)\\
{n_{C{O_2}}} = \dfrac{V}{{22,4}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
{n_{{H_2}O}} = \dfrac{m}{M} = \dfrac{{10,8}}{{18}} = 0,6mol\\
hh:C{H_4}(a\,mol),{C_2}{H_2}(b\,mol)\\
a + 2b = 0,4(1)\\
2a + b = 0,6(2)\\
\text{Từ (1) và (2)}\Rightarrow a = \dfrac{4}{{15}};b = \dfrac{1}{{15}}\\
{V_{C{H_4}}} = n \times 22,4 = \dfrac{4}{{15}} \times 22,4 = 5,973l\\
{V_{{C_2}{H_2}}} = n \times 22,4 = \dfrac{1}{{15}} \times 22,4 = 1,493l\\
\% C{H_4} = \dfrac{{5,973}}{{5,973 + 1,493}} \times 100\% = 80\% \\
\% {C_2}{H_2} = 100 - 80 = 20\%
\end{array}\)
