Em tham khảo nha:
\(\begin{array}{l}
1)\\
a)\\
etilen:{C_2}{H_4}\\
CTCT:C{H_2} = C{H_2}\\
{\rm{ax}}it\,{\rm{ax}}etic:{C_2}{H_4}{O_2}\\
CTCT:C{H_3}COOH\\
b)\\
{C_2}{H_5}OH + {O_2} \xrightarrow{\text{ lên men }} C{H_3}COOH + {H_2}O\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
C{H_3}COO{C_2}{H_5} + NaOH \to C{H_3}COONa + {C_2}{H_5}OH\\
2)\\
a)\,C{H_4} + C{l_2} \to C{H_3}Cl + HCl\\
b)\,nC{H_2} = C{H_2} \to {( - C{H_2} - C{H_2} - )_n}\\
c)\,2C{H_3}COOH + CaO \to {(C{H_3}COO)_2}Ca + {H_2}O\\
3)\\
a)\\
{C_2}{H_5}OH + 3{O_2} \xrightarrow{t^0} 2C{O_2} + 3{H_2}O\\
{n_{{C_2}{H_5}OH}} = \dfrac{{2,3}}{{46}} = 0,05\,mol\\
{n_{C{O_2}}} = 2{n_{{C_2}{H_5}OH}} = 0,1\,mol\\
{m_{C{O_2}}} = 0,1 \times 44 = 4,4g\\
b)\\
{n_{{H_2}O}} = 3{n_{{C_2}{H_5}OH}} = 0,15\,mol\\
{m_{{H_2}O}} = 0,15 \times 18 = 2,7g
\end{array}\)
