Đáp án:
b.$m = m_{CH_3COOH}+m_{CH_3COOC_2H_5} =0,15.60+0,3.88=35,4g$
c/ $\%m_{CH_3COOH}=25,42\%\\⇒\%m_{CH_3COOC_2H_5}=74,58\%$
Giải thích các bước giải:
a/ $CH_3COOH+NaOH\to CH_3COONa +H_2O(1)\\CH_3COOC_2H_5+NaOH\to CH_3COONa+C_2H_5OH(2)\\C_2H_5OH+Na\to C_2H_5ONa+\dfrac{1}{2}H_2(3)$
b. Gọi: $n_{CH_3COOH}=a; n_{CH_3COOC_2H_5}=b$
Theo (1),(2) ⇒ $a+b=n_{NaOH}=0,3.2,5=0,45\ mol(*)$
Theo (3) $n_{C_2H_5OH}=2.n_{H_2}=2.\dfrac{3,36}{22,4}=0,3\ mol$
Theo (2): $b = n_{CH_3COOC_2H_5} = n_{C_2H_5OH} =0,3$
Thay vào (*) ⇒ a = 0,15 mol
$m = m_{CH_3COOH}+m_{CH_3COOC_2H_5} =0,15.60+0,3.88=35,4g$
c/ $m_{CH_3COOH}=0,15.60=9g\\⇒\%m_{CH_3COOH}=\dfrac{9.100\%}{35,4}=25,42\%\\⇒\%m_{CH_3COOC_2H_5}=100-25,42=74,58\%$