Đáp án:
a) 62,5% và 37,5%
b) 24,08l
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}(1)\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}(2)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to A{g_2}{C_2} + 2N{H_4}N{O_3}(3)\\
{n_{B{r_2}}} = \dfrac{{104}}{{160}} = 0,65\,mol\\
{n_{A{g_2}{C_2}}} = \dfrac{{60}}{{240}} = 0,25\,mol\\
{n_{{C_2}{H_2}}} = {n_{A{g_2}{C_2}}} = 0,25\,mol\\
{n_{B{r_2}(1)}} = 2{n_{{C_2}{H_2}}} = 0,5\,mol\\
{n_{B{r_2}(2)}} = 0,65 - 0,5 = 0,15\,mol\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}(2)}} = 0,15\,mol\\
\% {n_{{C_2}{H_2}}} = \dfrac{{0,25}}{{0,25 + 0,15}} \times 100\% = 62,5\% \\
\% {n_{{C_2}{H_4}}} = 100 - 62,5 = 37,5\% \\
b)\\
2{C_2}{H_2} + 5{O_2} \xrightarrow{t^0} 4C{O_2} + 2{H_2}O\\
{C_2}{H_4} + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 2{H_2}O\\
{n_{{O_2}}} = 0,25 \times 2,5 + 0,15 \times 3 = 1,075\,mol\\
{V_{{O_2}}} = 1,075 \times 22,4 = 24,08l
\end{array}\)
