Câu 3:
Ta có: $n_{CH_3COOH}=n_{NaOH}=0,05(mol)$
$\Rightarrow n_{C_2H_5OH}=0,1(mol)$
$CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O$
Vì hiệu suất đạt 60% nên $n_{CH_3COOC_2H_5}=0,03(mol)$
$\Rightarrow m_{este}=2,64(g)$
Câu 2 :
\(2CH_3-CH_2-OH + 2Na \to 2CH_3-CH_2-ONa + H_2\\ 2CH_3COONa + Na_2CO_3 \to 2CH_3COONa + CO_2 + H_2O\\ Ca O+ 2CH_3COOH \to (CH_3COO)_2Ca + H_2O\)
Câu 3 :
a) CH3COOH + NaOH $\to$ CH3COONa + H2O
n CH3COOH = n NaOH = 0,05(mol)
=> m CH3COOH = 0,05.60 = 3(gam)
=> n C2H5OH = 7,6 -3 = 4,6(gam)
b) $CH_3COOH +C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
n C2H5OH = 4,6/46 = 0,1(mol) > n CH3COOH = 0,05 nên hiệu suất tính theo số mol CH3COOH
n CH3COOC2H5 = n CH3COOH pư = 0,05.60% = 0,03 mol
m CH3COOC2H5 = 0,03.88 = 2,64 gam
