Em tham khảo nha:

\(\begin{array}{l}
2)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to {C_2}A{g_2} + 2N{H_4}N{O_3}\\
{n_{{C_2}{H_2}}} = \dfrac{{7,84}}{{22,4}} = 0,35\,mol\\
{n_{{C_2}A{g_2}}} = {n_{{C_2}{H_2}}} = 0,35\,mol\\
{m_{{C_2}A{g_2}}} = 0,35 \times 240 = 84g\\
3)\\
a)\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{5,6}}{{22,4}} = 0,25\,mol\\
{n_{ancol}} = 2{n_{{H_2}}} = 0,5\,mol\\
{M_{ancol}} = \dfrac{{20,2}}{{0,5}} = 40,4g/mol\\
 \Rightarrow 14n + 18 = 40,4 \Rightarrow n = 1,6\\
 \Rightarrow CTHH:C{H_3}OH,{C_2}{H_5}OH\\
b)\\
hh:C{H_3}OH(a\,mol),{C_2}{H_5}OH(b\,mol)\\
\left\{ \begin{array}{l}
32a + 46b = 20,2\\
a + b = 0,5
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,3\\
\% {m_{C{H_3}OH}} = \dfrac{{0,2 \times 32}}{{20,2}} \times 100\%  = 31,68\% \\
\% {m_{{C_2}{H_5}OH}} = 100 - 31,68 = 68,32\% 
\end{array}\)