Question

Answer 1

a)

$n_{Na_2CO_3} = 0,05.1 = 0,05(mol)$

$2CH_3COOH + Na_2CO_3 \to 2CH_3COONa + CO_2 + H_2O$

$n_{CH_3COOH} = 2n_{Na_2CO_3} = 0,1(mol)$

$\%m_{CH_3COOH} = \dfrac{0,1.60}{8,3}.100\% = 72,3\%$

$\%m_{C_2H_5OH} = 100\% - 72,3\% = 27,7\%$

b) $n_{C_2H_5OH} = 0,05(mol)$

$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

Ta thấy : $n_{CH_3COOH} : 1 > n_{C_2H_5OH} : 1$ 

Suy ra : $n_{CH_3COOC_2H_5} = n_{C_2H_5OH\ pư} = 0,05.60\% = 0,03(mol)$
$m_{CH_3COOC_2H_5} = 0,03.88 = 2,64(gam)$