Em tham khảo nha:
\(\begin{array}{l}
4)\\
a)\\
C{H_4} + 2{O_2} \xrightarrow{t^0} C{O_2} + 2{H_2}O(1)\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0}4C{O_2} + 2{H_2}O(2)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}(3)\\
b)\\
{n_{B{r_2}}} = 0,1 \times 2 = 0,2\,mol\\
{n_{{C_2}{H_2}}} = \dfrac{{0,2}}{2} = 0,1\,mol\\
{n_{C{O_2}(2)}} = 2{n_{{C_2}{H_2}}} = 0,2\,mol\\
{V_{C{O_2}(2)}} = 0,2 \times 22,4 = 4,48l > 1,344l(?)\\
5)\\
a)\\
hh:\,C{H_4}(a\,mol),{C_2}{H_4}(b\,mol)\\
{n_{hh}} = \dfrac{{3,36}}{{22,4}} = 0,15\,mol\\
\left\{ \begin{array}{l}
16a + 28b = 3\\
a + b = 0,15
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,05\\
{V_{C{H_4}}} = 0,1 \times 22,4 = 2,24l\\
{V_{{C_2}{H_4}}} = 0,05 \times 22,4 = 1,12l\\
b)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{B{r_2}}} = {n_{{C_2}{H_4}}} = 0,05\,mol\\
{C_M}B{r_2} = \dfrac{{0,05}}{{0,05}} = 1M
\end{array}\)
