Đáp án:

$a)$

$CTPT:\ C_3H_6O_2$
$CTCT:\ CH_3-CH_2-COOH$

$b)\ m_{C_2H_5COOC_2H_5}=0,714(g)$

Giải thích các bước giải:

$a)$

$C_nH_{2n+1}COOH+NaOH\to C_nH_{2n+1}COONa+H_2O$

$n_{NaOH}=0,05.0,2=0,01(mol)$

$n_{C_nH_{2n+1}COOH}=n_{NaOH}=0,01(mol)$

$M_{C_nH_{2n+1}COOH}=\dfrac{0,74}{0,01}=74(g/mol)$

Mà: $14n+46=74(g/mol)$

$\to n=2$

Vậy axit là: $C_2H_5COOH$

$CTPT:\ C_3H_6O_2$

$CTCT:\ CH_3-CH_2-COOH$

$b)$

$n_{C_2H_5COOH}=\dfrac{0,74}{74}=0,01(mol)$

$C_2H_5COOH+C_2H_5OH\to C_2H_5COOC_2H_5+H_2O$

$n_{C_2H_5COOC_2H_5}=n_{C_2H_5COOH}=0,01(mol)$

$m_{C_2H_5COOC_2H_5}=0,01.102.70\%=0,714(g)$

Em tham khảo nha:

\(\begin{array}{l}
a)\\
{C_n}{H_{2n + 1}}COOH + NaOH \to {C_n}{H_{2n + 1}}COONa + {H_2}O\\
{n_{NaOH}} = 0,05 \times 0,2 = 0,01\,mol\\
{n_{{C_n}{H_{2n + 1}}COOH}} = {n_{NaOH}} = 0,01\,mol\\
{M_{{C_n}{H_{2n + 1}}COOH}} = \dfrac{{0,74}}{{0,01}} = 74g/mol\\
 \Rightarrow 14n + 46 = 74 \Rightarrow n = 2\\
CTPT:{C_3}{H_6}{O_2}\\
CTCT:C{H_3} - C{H_2} - COOH\\
b)\\
C{H_3} - C{H_2} - COOH + {C_2}{H_5}OH \to C{H_3} - C{H_2} - COO{C_2}{H_5} + {H_2}O\\
{n_{C{H_3}C{H_2}COO{C_2}{H_5}}} = {n_{C{H_3} - C{H_2} - COOH}} = 0,01\,mol\\
H = 70\%  \Rightarrow {m_{C{H_3}C{H_2}COO{C_2}{H_5}}} = 0,01 \times 102 \times 70\%  = 0,714g
\end{array}\)