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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%
\(n_{CO2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
Pt : \(C_6H_{12}O_6\xrightarrow[30-35^oC]{Menrượu}2C_2H_5OH+2CO_2\)
0,5 0,5
a) \(m_{C2H5OH}=0,5.46=23\left(g\right)\)
b) Pt : \(C_2H_5OH+O_2\xrightarrow[]{Mengiấm}CH_3COOH+H_2O\)
0,5 0,5
\(m_{CH3COOH\left(lt\right)}=0,5.60=30\left(g\right)\)
⇒ \(m_{CH3COOH\left(tt\right)}=30.80\%=24\left(g\right)\)
Chúc bạn học tốt
\(C_6H_{12}O_6\underrightarrow{t^o}2C_2H_5OH+2CO_2\uparrow\)(xt : men rượu )
0,5 0,5
\(n_{CO_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
\(m_{C_2H_5OH}=0,5.46=23\left(g\right)\)
\(C_2H_5OH+O_2\underrightarrow{t^o}CH_3COOH+H_2O\) (men giấm )
0,5 0,5
\(m_{CH_3COOH}=0,5.60=30\left(g\right)\)
\(m_{CH_3COOHtt}=30.80\%=24\left(g\right)\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
1 0,5 ( mol )
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)
1 1 ( mol )
\(m_{CH_3COOH}=1.60.80\%=48g\)
có: nH2= \(\frac{1,96}{22,4}\)= 0,0875( mol)
nKOH= 2. 0,05= 0,1( mol)
PTPU
2C2H5OH+ 2Na\(\rightarrow\) 2C2H5ONa+ H2\(\uparrow\) ( 1)
2CH3COOH+ 2Na\(\rightarrow\) 2CH3COONa+ H2\(\uparrow\) ( 2)
.0,1.......................................................0,05.. ( mol)
CH3COOH+ KOH\(\rightarrow\) CH3COOK+ H2O ( 3)
..0,1................0,1.................................... mol
có: \(\sum\)nH2= nH2( 1)+ nH2( 2)= 0,0875( mol)
\(\Rightarrow\) nH2( 1)= 0,0875- 0,05= 0,0375( mol)
theo pt( 1) có: nC2H5OH= 2nH2= 2. 0,0375= 0,075( mol)
\(\Rightarrow\) mhh= 0,1. 60+ 0,075. 46= 9,45( g)
\(\Rightarrow\) %mC2H5OH= \(\frac{0,075.46}{9,45}\). 100%= 36,5%
%mCH3COOH= 100%- 36,5%= 63,5%
b/
C2H5OH+ CH3COOH<=( H2SO4 đặc; to)=> CH3COOC2H5+ H2O ( 4)
xét tỉ lệ: \(\frac{0,075}{1}\)< \(\frac{0,1}{1}\)
\(\Rightarrow\) C2H5OH hết, CH3COOH dư
theo PTPU( 4) có: nCH3COOC2H5= nC2H5OH= 0,075( mol)
mặt khác: nCH3COOC2H5( thực tế)= \(\frac{5,28}{88}\)= 0,06( mol)
\(\Rightarrow\) H= \(\frac{0,06}{0,075}\). 100%= 80%
\(a,n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5\left(mol\right)\)
PTHH: C6H12O6 --men rượu--> 2C2H5OH + 2CO2
2,5-------------------------->5
C2H5OH + O2 --men giấm--> CH3COOH + H2O
5------------------------------------>5
\(b,m_{C_2H_5OH}=5.46=230\left(g\right)\)
\(c,m_{CH_3COOH}=5.80\%.60=240\left(g\right)\)
a)\(n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5mol\)
\(C_6H_{12}O_6\underrightarrow{menrượu}2C_2H_5OH+2CO_2\)
2,5 5
b)\(m_{C_2H_5OH}=5\cdot46=230g\)
c)\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)
5 5
Thực tế: \(n_{CH_3COOH}=5\cdot80\%=4mol\)
\(m_{CH_3COOH}=4\cdot60=240g\)