Đáp án:
\(\% {V_{C{H_4}}} = 37,5\% \)
Giải thích các bước giải:
\(\begin{array}{l}
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to A{g_2}{C_2} + 2N{H_4}N{O_3}\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
{n_{A{g_2}{C_2}}} = \dfrac{{24}}{{240}} = 0,1\,mol\\
{n_{{C_2}{H_2}}} = {n_{A{g_2}{C_2}}} = 0,1\,mol\\
{n_{{C_2}{H_4}}} = \dfrac{{4,2}}{{28}} = 0,15\,mol\\
{V_{C{H_4}}} = 8,96 - 0,15 \times 22,4 - 0,1 \times 22,4 = 3,36l\\
\% {V_{C{H_4}}} = \dfrac{{3,36}}{{8,96}} \times 100\% = 37,5\%
\end{array}\)