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Ta có : nCH3COOH = 0,2(mol) ; nC2H5OH = 0,03(mol)
PTHH :
\(CH3COOH+C2H5OH-^{H2SO4,đặc,t0}->CH3COOC2H5+H2O\)
Ta có : nCH3COOH = 0,2 > nC2H5OH = 0,03 => nCH3COOH dư
=> nCH3COOC2H5 = nC2H5OH = 0,03
=> m(etyl axetat) = 0,03.88 = 2,64(g)
=> m(etyl axtat)(thực tế) = \(\dfrac{2,64.75}{100}=1,98\left(g\right)\)
\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,05<--------------------------------------0,05
=> \(m_{CH_3COOH\left(lý.thuyết\right)}=0,05.60=3\left(g\right)\)
=> \(m_{CH_3COOH\left(tt\right)}=\dfrac{3.100}{60}=5\left(g\right)\)
1,
- Xét phần 2:
\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)
PTHH: CH3COOH + C2H5OH \(\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}\) CH3COOC2H5 + H2O
LTL: 0,5a < 0,5b (do a < b) => C2H5OH dư
Theo pthh: \(n_{CH_3COOH\left(pư\right)}=n_{CH_3COOC_2H_5}=0,05\left(mol\right)\)
Mà H = 50%
=> \(n_{CH_3COOH\left(bđ\right)}=\dfrac{0,05}{50\%}=0,1\left(mol\right)\)
Xét phần 2:
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH:
2CH3COOH + 2Na ---> 2CH3COONa + H2
0,1--------------------------------------------->0,05
2C2H5OH + 2Na ---> 2C2H5ONa + H2
0,4<----------------------------------------0,2
=> Trong X có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=0,1.2=0,2\left(mol\right)\\n_{C_2H_5OH}=0,4.2=0,8\left(mol\right)\end{matrix}\right.\)
2, Gọi \(\left\{{}\begin{matrix}n_{CH_3COOH}=a\left(mol\right)\\n_{C_2H_5OH}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(n_{H_2O}=\dfrac{23,4}{18}=1,3\left(mol\right)\)
PTHH:
CH3COOH + 2O2 --to--> 2CO2 + 2H2O
a------------------------------------------>2a
C2H5OH + 3O2 --to--> 2CO2 + 3H2O
b-------------------------------------->3b
=> Hệ pt \(\left\{{}\begin{matrix}60a+46b=25,8\\2a+3b=1,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\left(TM\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
LTL: 0,2 < 0,3 => Rượu dư
Theo pthh: \(n_{CH_3COOH\left(pư\right)}=n_{CH_3COOC_2H_5}=\dfrac{14,08}{88}=0,16\left(mol\right)\)
=> \(H=\dfrac{0,16}{0,2}.100\%=80\%\)
CH3COOH=0,15 mol
C2H5OH=0,1 mol
C2H5OH+CH3COOH->CH3COOC2H5+H2O
0,075-------------------------------0,075
=>CH3COOH dư
n este =0,075 mol
=>H=\(\dfrac{0,075}{0,1}\)100=75%
\(n_{CH_3COOH}=\dfrac{9}{60}=0,15\left(mol\right)\\ n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\\ CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\\ LTL:\dfrac{0,15}{1}>\dfrac{0,1}{1}\\ \Rightarrow TínhtheosốmolC_2H_5OH\\\Rightarrow n_{CH_3COOC_2H_5\left(lt\right)}=n_{C_2H_5OH}=0,1\left(mol\right)\\ n_{CH_3COOC_2H_5\left(tt\right)}=\dfrac{6,6}{88}=0,075\left(mol\right)\\ \Rightarrow H=\dfrac{0,075}{0,1}.100=75\%\)
CH3COOH + NaOH $\to$ CH3COONa + H2O
n CH3COOH = n NaOH = 0,05(mol)
=> n C2H5OH = (7,6 - 0,05.60)/46 = 0,1(mol)
\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)
n CH3COOH = 0,05 < n C2H5OH = 0,1 nên hiệu suất tính theo số mol CH3COOH
n CH3COOC2H5 = n CH3COOH pư = 0,05.60% = 0,03 mol
=> m este = 0,03.88 = 2,64 gam