Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{C{H_4}}} = 50\% \\
\% {V_{{C_2}{H_4}}} = 50\% \\
c)\\
{m_{B{r_2}}} = 32g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
C{H_4} + 2{O_2} \to C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \to 2C{O_2} + 2{H_2}O\\
b)\\
{n_{{H_2}O}} = \dfrac{{14,4}}{{18}} = 0,8mol\\
{n_{hh}} = \dfrac{{8,96}}{{22,4}} = 0,4mol\\
hh:C{H_4}(a\,mol);{C_2}{H_4}(b\,mol)\\
\left\{ \begin{array}{l}
a + b = 0,4\\
2a + 2b = 0,8
\end{array} \right.\\
\Rightarrow a = 0,2;b = 0,2\\
\% {V_{C{H_4}}} = \dfrac{{0,2 \times 22,4}}{{8,96}} = 50\% \\
\% {V_{{C_2}{H_4}}} = 100 - 50 = 50\% \\
c)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = {n_{C{H_4}}} = 0,2mol\\
{m_{B{r_2}}} = 0,2 \times 160 = 32g
\end{array}\)
