Đáp án:

\(\% {m_{{C_2}{H_4}}} = 43\% ;\% {m_{{C_2}{H_2}}} = 20\% \)

\(\% {m_{C{H_4}}} = 37\% \)

Giải thích các bước giải:

 Các phản ứng xảy ra:

\({C_2}{H_2} + 2{\text{A}}gN{O_3} + 2N{H_3}\xrightarrow{{}}A{g_2}{C_2} + 2N{H_4}N{O_3}\)

\({C_2}{H_4} + B{{\text{r}}_2}\xrightarrow{{}}{C_2}{H_4}B{{\text{r}}_2}\)

Ta có:

\({n_{A{g_2}{C_2}}} = \frac{{24}}{{108.2 + 12.2}} = 0,1{\text{ mol = }}{{\text{n}}_{{C_2}{H_2}}}\)

Khối lượng bình 2 tăng là khối lượng của \(C_2H_4\)

\( \to {m_{{C_2}{H_4}}} = 5,6{\text{ gam}}\)

\( \to {n_{{C_2}{H_4}}} = \frac{{5,6}}{{28}} = 0,2{\text{ mol}}\)

\({m_{{C_2}{H_2}}} = 0,1.26 = 2,6{\text{ gam}}\)

\({n_{hh}} = \frac{{13,44}}{{22,4}} = 0,6{\text{ mol}} \to {{\text{n}}_{C{H_4}}} = 0,6 - 0,2 - 0,1 = 0,3{\text{ mol}}\)

\( \to {m_{C{H_4}}} = 0,3.16 = 4,8{\text{ gam}}\)

\( \to \% {m_{{C_2}{H_4}}} = \frac{{5,6}}{{5,6 + 2,6 + 4,8}} = 43\% ;\% {m_{{C_2}{H_2}}} = \frac{{2,6}}{{5,6 + 2,6 + 4,8}} = 20\% \)

\( \to \% {m_{C{H_4}}} = 37\% \)

$n_{hh}=\dfrac{13,44}{22,4}=0,6\ (mol)$

Bình 1:

$C_2H_2+2AgNO_3+2NH_3\to C_2Ag_2+2NH_4NO_3$

$n_{C_2Ag_2}=\dfrac{24}{240}=0,1\ (mol)$

Bình 2:

$m_{\text{bình tăng}}=m_{C_2H_4}=5,6\ (g)$

$\to n_{C_2H_4}=\dfrac{5,6}{28}=0,2\ (mol)$

$\to n_{CH_4}=0,6-0,1-0,2=0,3\ (mol)$

$m_{hh}=0,1.26+0,2.28+0,3.16=13\ (g)$

$\%m_{CH_4}=\dfrac{0,3.16}{13}.100\%=36,92\%$

$\%m_{C_2H_4}=\dfrac{0,2.28}{13}.100\%=43,08\%$

$\%m_{C_2H_2}=100\%-36,92\%-43,08\%=20\%$