Em tham khảo nha:
\(\begin{array}{l}
1)\\
{C_2}{H_2} + {H_2} \xrightarrow{PbCl_2,t^0} {C_2}{H_4}\\
{C_2}{H_4} + {H_2}O \xrightarrow{xt,t^0} {C_2}{H_5}OH\\
{C_2}{H_5}OH + CuO \to C{H_3}CHO + Cu + {H_2}O\\
2)\\
a)\\
{C_2}{H_2} + 2AgN{O_3} + 2N{H_3} \to A{g_2}{C_2} + 2N{H_4}N{O_3}\\
C{H_3}CHO + 2AgN{O_3} + 3N{H_3} + {H_2}O \to C{H_3}COON{H_4} + 2N{H_4}N{O_3} + 2Ag\\
b)\\
hh:{C_2}{H_2}(a\,mol),C{H_3}CHO(b\,mol)\\
\left\{ \begin{array}{l}
26a + 44b = 4,8\\
240a + 108 \times 2b = 34,8
\end{array} \right.\\
\Rightarrow a = 0,1;b = 0,05\\
\% {m_{{C_2}{H_2}}} = \dfrac{{0,1 \times 26}}{{4,8}} \times 100\% = 54,17\% \\
\% {m_{C{H_3}CHO}} = 100 - 54,17 = 45,83\%
\end{array}\)
