Đáp án:
A
Giải thích các bước giải:
\(\begin{array}{l}
hh:{C_2}{H_5}OH(a\,mol),C{H_3}COOH(b\,mol),HCOOC{H_3}(c\,mol)\\
{P_1}:\\
2{C_2}{H_5}OH + 2Na \to 2{C_2}{H_5}OH + {H_2}\\
2C{H_3}COOH + 2Na \to 2C{H_3}COONa + {H_2}\\
{n_{{H_2}}} = \dfrac{{4,48}}{{22,4}} = 0,2\,mol\\
\Rightarrow \dfrac{a}{3} \times \dfrac{1}{2} + \dfrac{b}{3} \times \dfrac{1}{2} = 0,2 \Rightarrow a + b = 1,2(1)\\
{P_2}:\\
C{H_3}COOH + NaOH \to C{H_3}COONa + {H_2}O\\
HCOOC{H_3} + NaOH \to HCOONa + C{H_3}OH\\
{n_{NaOH}} = 0,2 \times 1 = 0,2\,mol\\
\Rightarrow \dfrac{b}{3} + \dfrac{c}{3} = 0,2 \Rightarrow b + c = 0,6(2)\\
{P_3}:\\
{C_2}{H_5}OH +3 {O_2} \to 2C{O_2} + 3{H_2}O\\
C{H_3}COOH +2 {O_2} \to 2C{O_2} + 2{H_2}O\\
HCOOC{H_3} + 2{O_2} \to 2C{O_2} +2 {H_2}O\\
{n_{C{O_2}}} = \dfrac{{39,6}}{{44}} = 0,9\,mol\\
\Rightarrow \dfrac{a}{3} \times 2 + \dfrac{b}{3} \times 2 + \dfrac{c}{3} \times 2 = 0,9 \Rightarrow a + b + c = 1,35(3)\\
(1),(2),(3) \Rightarrow a = 0,75;b = 0,45;c = 0,15\\
m = 0,75 \times 46 + 0,45 \times 60 + 0,15 \times 60 = 70,5g
\end{array}\)
