Đáp án:
47 B
50 D
Giải thích các bước giải:
\(\begin{array}{l}
47)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
hh:C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
{n_{{O_2}}} = \dfrac{{12,8}}{{32}} = 0,4\,mol\\
{n_{hh}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\left\{ \begin{array}{l}
a + b = 0,15\\
2a + 3b = 0,6
\end{array} \right.\\
\Rightarrow a = 0,05;b = 0,1\,mol\\
{V_{C{H_4}}} = 0,05 \times 22,4 = 1,12l\\
{V_{{C_2}{H_4}}} = 0,1 \times 22,4 = 2,24l\\
50)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{n_{hh}} = \dfrac{{0,56}}{{22,4}} = 0,025\,mol\\
hh:{C_2}{H_4}(a\,mol),{C_2}{H_2}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,025\\
28a + 26b = 0,675
\end{array} \right.\\
\Rightarrow a = b = 0,0125\\
\% {V_{{C_2}{H_4}}} = \dfrac{{0,0125}}{{0,025}} \times 100\% = 50\% \\
\% {V_{{C_2}{H_2}}} = 100 - 50 = 50\%
\end{array}\)
