Đáp án:
50% 25% và 25%
Giải thích các bước giải:
\(\begin{array}{l}
2{C_2}{H_6} + 7{O_2} \xrightarrow{t^0} 4C{O_2} + 6{H_2}O\\
2{C_3}{H_6} + 9{O_2} \xrightarrow{t^0} 6C{O_2} + 6{H_2}O\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{C_3}{H_6} + B{r_2} \to {C_3}{H_6}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_2}\\
{n_{C{O_2}}} = \dfrac{{10,08}}{{22,4}} = 0,45\,mol\\
{n_{B{r_2}}} = \dfrac{{24}}{{160}} = 0,15\,mol\\
hh:{C_2}{H_6}(a\,mol),{C_3}{H_6}(b\,mol),{C_2}{H_2}(c\,mol)\\
\left\{ \begin{array}{l}
30a + 42b + 26c = 6,4\\
2a + 3b + 2c = 0,45\\
b + 2c = 0,15
\end{array} \right.\\
\Rightarrow a = 0,1;b = c = 0,05\\
\% {V_{{C_2}{H_6}}} = \dfrac{{0,1}}{{0,1 + 0,05 + 0,01}} \times 100\% = 50\% \\
\% {V_{{C_3}{H_6}}} = \% {V_{{C_2}{H_2}}} = \dfrac{{0,1}}{{0,1 + 0,05 + 0,01}} \times 100\% = 25\%
\end{array}\)
