\(3CH_3COOH + C_3H_5(OH)_3 \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons (CH_3COO)_3C_3H_5 + 3H_2O\)
Ta thấy :
$n_{glixerol} : 1 = 0,1 < n_{CH_3COOH} : 3 = 0,05$ nên hiệu suất tính theo axit
$n_{este} = \dfrac{1}{3}n_{CH_3COOH\ pư} = \dfrac{1}{3}.0,15.60\% = 0,03(mol)$
$m = 0,03.218 = 6,54(gam)$
\(2CH_3COOH+C_3H_5\left(OH\right)_3\underrightarrow{^{^{H^+}}}\left(CH_3COO\right)_3C_3H_5+3H_2O\)
Lập tỉ lệ : \(\dfrac{0.15}{2}< \dfrac{0.1}{1}\)
Tính theo axit axetic
\(m_{X\left(tt\right)}=0.075\cdot218\cdot60\%=9.81\left(g\right)\)