Đáp án:
\(\begin{array}{l}
\% {C_2}{H_2} = 48,15\% \\
\% {C_2}{H_4} = 51,85\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_2} + 2B{r_2} \to {C_2}{H_2}B{r_4}\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to 2N{H_4}N{O_3} + A{g_2}{C_2}\\
{n_{B{r_2}}} = \dfrac{{4,8}}{{160}} = 0,03mol\\
{n_{A{g_2}{C_2}}} = \dfrac{{2,4}}{{240}} = 0,01mol\\
{n_{{C_2}{H_2}}} = {n_{A{g_2}{C_2}}} = 0,01mol\\
{m_{{C_2}{H_2}}} = 0,01 \times 26 = 0,26g\\
{n_{{C_2}{H_4}}} = {n_{B{r_2}}} - 2{n_{{C_2}{H_2}}} = 0,03 - 0,02 = 0,01mol\\
{m_{{C_2}{H_4}}} = 0,01 \times 28 = 0,28g\\
\% {C_2}{H_2} = \dfrac{{0,26}}{{0,26 + 0,28}} \times 100\% = 48,15\% \\
\% {C_2}{H_4} = 100 - 48,15 = 51,85\%
\end{array}\)
