Đáp án:
\(\begin{array}{l}
a)\\
\% {V_{C{H_4}}} = 16,67\% \\
\% {V_{{C_2}{H_4}}} = 50\% \\
\% {V_{{C_2}{H_2}}} = 33,33\% \\
b)\\
{m_{A{g_2}{C_2}}} = 24g
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to A{g_2}{C_2} + 2N{H_4}N{O_3}\\
X:C{H_4},{C_2}{H_4}\\
{V_{{C_2}{H_2}}} = 6,72 - 4,48 = 2,24l\\
{C_2}{H_4} + B{r_2} \to {C_2}{H_4}B{r_2}\\
Y:C{H_4}\\
{V_{C{H_4}}} = {V_Y} = 1,12l\\
{V_{{C_2}{H_4}}} = 4,48 - 1,12 = 3,36l\\
\% {V_{C{H_4}}} = \dfrac{{1,12}}{{6,72}} \times 100\% = 16,67\% \\
\% {V_{{C_2}{H_4}}} = \dfrac{{3,36}}{{6,72}} \times 100\% = 50\% \\
\% {V_{{C_2}{H_2}}} = 100 - 50 - 16,67 = 33,33\% \\
b)\\
{n_{{C_2}{H_2}}} = \dfrac{{2,24}}{{22,4}} = 0,1\,mol\\
{n_{A{g_2}{C_2}}} = {n_{{C_2}{H_2}}} = 0,1\,mol\\
{m_{A{g_2}{C_2}}} = 0,1 \times 240 = 24g
\end{array}\)