chúc bn hc tốt cho mk xin ctlhn và 5* ạ

Em tham khảo nha :

\(\begin{array}{l}
1)\\
2{C_n}{H_{2n + 1}}OH + 2Na \to 2{C_n}{H_{2n + 1}}ONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{1,12}}{{22,4}} = 0,05mol\\
{n_A} = 2{n_{{H_2}}} = 0,1mol\\
{M_A} = \dfrac{6}{{0,1}} = 60dvC\\
n = \dfrac{{60 - 18}}{{14}} = 3\\
CTHH:{C_3}{H_7}OH\\
CTCT:\\
C{H_3} - C{H_2} - C{H_2}OH:propan - 1 - ol\\
C{H_3} - CH(OH) - C{H_3}:propan - 2 - ol\\
2)\\
hh:HCOOH(a\,mol);C{H_3}COOH(b\,mol)\\
\left\{ \begin{array}{l}
a - b = 0\\
46a + 60b = 21,2
\end{array} \right.\\
 \Rightarrow a = 0,2;b = 0,2\\
{n_{{C_2}{H_5}OH}} = \dfrac{{23}}{{46}} = 0,5mol\\
HCOOH + {C_2}{H_5}OH \to HCOO{C_2}{H_5} + {H_2}O\\
C{H_3}COOH + {C_2}{H_5}OH \to C{H_3}COO{C_2}{H_5} + {H_2}O\\
\dfrac{{0,2 \times 2}}{1} < \dfrac{{0,5}}{1} \Rightarrow {C_2}{H_5}OH\text{ dư}\\
{n_{HCOO{C_2}{H_5}}} = {n_{HCOOH}} = 0,2mol\\
{m_{HCOO{C_2}{H_5}}} = 0,2 \times 62 = 12,4g\\
{n_{C{H_3}COO{C_2}{H_5}}} = {n_{C{H_3}COOH}} = 0,2mol\\
{m_{C{H_3}COO{C_2}{H_5}}} = 0,2 \times 88 = 17,6g\\
{m_{este}} = \dfrac{{(12,4 + 17,6) \times 80}}{{100}} = 24g
\end{array}\)