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Ta có:
\(V_{C2H5OH}=4,6.14\%=0,644\left(l\right)=644\left(ml\right)\)
\(\Rightarrow m_{C2H5OH}=0,8.644=515,2\left(g\right)\)
\(\Rightarrow n_{C2H5OH}=\frac{515,2}{46}=11,2\left(mol\right)\)
\(V_{H2O}=4,6-0,644=3,956\left(l\right)=3956\left(ml\right)\)
\(\Rightarrow m_{H2O}=3956\left(g\right)\)
Mà H = 30%
\(\Rightarrow n_{C2H5OH\left(pư\right)}=11,2.30\%=3,36\left(mol\right)\)
\(C_2H_5OH+O_2\underrightarrow{^{men.giam}}CH_3COOH+H_2O\)
\(\Rightarrow n_{O2}=n_{CH3COOH}=3,36\left(mol\right)\)
\(m_{dd\left(spu\right)}=515,2+3956+3,36.32=4578,72\left(g\right)\)
\(\Rightarrow C\%_{CH3COOH}=\frac{3,36.60}{4578,72}.100\%=4,4\%\)
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ V_{C_2H_5OH}=25.4\%=1\left(l\right)=1000\left(ml\right)\\ m_{C_2H_5OH}=1000.0,8=800\left(g\right)\\ m_{CH_3COOH\left(LT\right)}=\dfrac{800.60}{46}=\dfrac{48000}{46}\left(g\right)\\ m_{CH_3COOH\left(TT\right)}=\dfrac{48000}{46}:92\%=1134,2155\left(gam\right)\\ m_{ddCH_3COOH}=1135,2155:5\%=22684,31\left(g\right)\)
V C 2 H 5 OH = 50.4/100 = 2l
→ m C 2 H 5 OH = 2.1000.0,8 = 1600g
Phương trình hóa học :
C 2 H 5 OH + O 2 → CH 3 COOH + H 2 O
46 gam 60 gam
1600 gam x
x = 1600x60/46
Vì hiệu suất đạt 80% → m CH 3 COOH = 1600.60.80/(46.100) = 1669,6g
→ m giấm = 1669,6/5 x 100 = 33392 (gam) = 33,392 kg
a.\(V_{C_2H_5OH}=\dfrac{10.8}{100}=0,8ml\)
\(m_{C_2H_5OH}=0,8.0,8=1,6g\)
\(n_{C_2H_5OH}=\dfrac{1,6}{46}=0,034mol\)
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)
0,034 0,034 ( mol )
\(m_{CH_3COOH}=0,034.60.80\%=1,632g\)
b.\(m_{dd_{CH_3COOH}}=\dfrac{1,632}{5\%}=32,64g\)
a) n Zn = 6,5/65 = 0,1(mol)
Zn + 2CH3COOH $\to$ (CH3COO)2Zn + H2
Theo PTHH :
n CH3COOH = 2n Zn =0,2(mol)
C% CH3COOH = 0,2.60/200 .100% = 6%
b) n H2 = n Zn = 0,1(mol)
=> m dd sau pư = 6,5 + 200 - 0,1.2 = 206,3 gam
Theo PTHH : n (CH3COO)2Zn = n Zn = 0,1(mol)
=> C% (CH3COO)2Zn = 0,1.183/206,3 .100% = 8,87%
c)
C2H5OH + O2 $\xrightarrow{men\ giấm}$ CH3COOH + H2O
n C2H5OH pư = n CH3COOH = 0,2(mol)
=> m C2H5OH cần dùng = 0,2.46/80% = 11,5 gam
a) nZn=0,1(mol)
PTHH: Zn + 2 CH3COOH -> (CH3COO)2Zn + H2
0,1_______0,2_________0,1_____________0,1(mol)
mCH3COOH=0,2.60=12(g)
=> C%ddCH3COOH=(12/200).100=6%
b) mdd(CH3COO)2Zn= 6,5+200-0,1.2=206,3(g)
m(CH3COO)2Zn= 183 x 0,1=18,3(g)
=>C%dd(CH3COO)2Zn= (18,3/206,3).100=8,871%
c) C2H5OH + O2 -men giấm-> CH3COOH + H2O
nC2H5OH(LT)=nCH3COOH=0,2(mol)
=> nC2H5OH(TT)=0,2 : 80%= 0,25(mol)
=>mC2H5OH=0,25 x 46= 11,5(g)
a)
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
V rượu = 57,5.12/100 = 6,9(lít) = 6900(cm3)
=> m rượu = 6900.0,8 = 5520(gam)
Theo PTHH :
n CH3COOH = n C2H5OH = 5520/46 = 120(mol)
m CH3COOH = 120.60 = 7200(gam)
b)
m dd giấm = 7200/4% = 180 000(gam)
\(V_r=57.5\cdot0.12=6.9\left(l\right)\)
\(m_{C_2H_5OH}=6.9\cdot0.8=5.52\left(g\right)\)
\(n_{C_2H_5OH}=\dfrac{5.52}{46}=0.12\left(mol\right)\)
\(n_{C_2H_5OH\left(pư\right)}=0.12\cdot92\%=0.1104\left(mol\right)\)
\(C_2H_5OH+O_2\underrightarrow{mg}CH_3COOH+H_2O\)
\(0.1104........................0.1104\)
\(m_{dd_{CH_3COOH}}=\dfrac{0.1104\cdot60}{4\%}=165.6\left(g\right)\)
a) n glucozo = 54/180 = 0,3(mol)
n glucozo pư = 0,3.80% = 0,24(mol)
$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 +2 C_2H_5OH$
n C2H5OH = 2n glucozo = 0,48(mol)
m C2H5OH = 0,48.46 = 22,08(gam)
b)
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = n C2H5OH = 0,48(mol)
C% CH3COOH = 0,48.60/500 .100% = 5,76%
\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25mol\)
\(C_6H_{12}O_6\underrightarrow{lênmen}2C_2H_5OH+2CO_2\)
0,25 0,5 0,5
\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)
0,5 0,5
\(C_{M_{CH_3COOH}}=\dfrac{0,5}{1}=0,5M\)
\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25\left(mol\right)\)
PTHH:
C6H12O6 --men rượu--> 2CO2 + 2C2H5OH
0,25-------------------------------------->0,5
C2H5OH + O2 --men giấm--> CH3COOH + H2O
0,5------------------------------------->0,5
\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{1}=0,5M\)
\(\left(C_6H_{10}O_5\right)_n+nH_2O\underrightarrow{^{axit}}nC_6H_{12}O_6\)
\(C_6H_{12}O_6\underrightarrow{^{men}}2C_2H_5OH+2CO_2\)
Ta có:
\(H=80\%.81\%=65,8\%\)
Sơ đồ : \(C_6H_{10}O_5\rightarrow C_6H_{12}O_6\rightarrow2C_2H_5OH\)
\(n_{C6H10O5}=\frac{1,5}{162}\Rightarrow n_{ancol\left(LT\right)}=\frac{1,5}{162}.2=\frac{3}{162}\left(mol\right)\)
\(\Rightarrow n_{ancol\left(tt\right)}=\frac{3}{162}.64,8\%=0,012\left(mol\right)\)
\(\Rightarrow m_{ancol}=0,012.46=0,552\left(g\right)\)