Đáp án:
\(\% {V_{C{H_4}}} = \% {V_{{C_2}{H_6}}} = 50\% \)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
2{C_2}{H_6} + 7{O_2} \xrightarrow{t^0} 4C{O_2} + 6{H_2}O\\
C{O_2} + Ca{(OH)_2} \to CaC{O_3} + {H_2}O\\
b)\\
{n_{CaC{O_3}}} = \dfrac{{30}}{{100}} = 0,3\,mol\\
{n_{C{O_2}}} = {n_{CaC{O_3}}} = 0,3\,mol\\
{n_X} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
hh\,X:C{H_4}(a\,mol),{C_2}{H_6}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,2\\
a + 2b = 0,3
\end{array} \right.\\
\Rightarrow a = b = 0,1\\
\% {V_{C{H_4}}} = \% {V_{{C_2}{H_6}}} = \dfrac{{0,1}}{{0,1 + 0,1}} \times 100\% = 50\%
\end{array}\)