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a)
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
b)
n CH3COOC2H5 = n C2H5OH = 9,2/46 = 0,2(mol)
=> m este = 0,2.88 = 17,6 gam
c)
n este = 8,8/88 = 0,1(mol)
=> n C2H5OH = n CH3COOH = 0,1/60% = 1/6 mol
=> m C2H5OH = 46 . 1/6 = 7,67(gam) ; m CH3COOH = 60 . 1/6 = 10(gam)
nC2H5OH = 8.05/46 = 0.175 (mol)
nCH3COOH = 36/60 = 0.6 (mol)
nCH3COOC2H5 = 12.32/88 = 0.14 (mol)
C2H5OH + CH3COOH <-H2SO4đ,t0-> CH3COOC2H5 + H2O
1.......................1
0.175................0.6
LTL : 0.175/1 < 0.6/1
=> CH3COOH dư
mCH3COOH (dư) = ( 0.6 - 0.175) * 60 = 25.5 (g)
nCH3COOC2H5 = nC2H5OH = 0.175 (mol)
H% = 0.14/0.175 * 100% = 80%
\(n_{C_2H_4}=\dfrac{1,12}{22,4}=0,05mol\)
\(C_2H_4+H_2O\rightarrow\left(t^o,H_2SO_4\right)C_2H_5OH\)
0,05 0,05 ( mol )
\(C_2H_5OH+CH_3COOH\rightarrow\left(t^o,H_2SO_4\right)CH_3COOC_2H_5+H_2O\)
0,05 0,05 ( mol )
\(m_{CH_3COOC_2H_5}=0,05.88=4,4g\)
\(a) C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ b) n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{9,2}{46} = 0,2(mol)\\ m_{CH_3COOH} = 0,2.60 = 12(gam)\\ c) n_{CH_3COOC_2H_5} = n_{C_2H_5OH} = 0,2(mol)\\ m_{CH_3COOC_2H_5} = 0,2.88 = 17,6(gam)\)
\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,05<--------------------------------------0,05
=> \(m_{CH_3COOH\left(lý.thuyết\right)}=0,05.60=3\left(g\right)\)
=> \(m_{CH_3COOH\left(tt\right)}=\dfrac{3.100}{60}=5\left(g\right)\)
A tác dụng với NaHCO3 cho khí CO2 → A: axit CH3COOH
BTKL: m + mO2 = mCO2 + mH2O => m = 1,8
=> nCH3COOH = 0,03
CH3COOH + C2H5OH → CH3COOC2H5 + H2O
0,03 0,02 0,0125
=> H = 62,5%