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\(n_{CO2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)
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
\(m_{CH_3COOH}=0,5.60=30\left(g\right)\)
\(m_{CH_3COOHtt}=30.80\%=24\left(g\right)\)
\(a,n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5\left(mol\right)\)
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: \(n_{CH_3COOH}=5\cdot80\%=4mol\)
\(m_{CH_3COOH}=4\cdot60=240g\)
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%
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, \(V_{C_2H_5OH}=\dfrac{10.9}{100}=0,9\left(l\right)=900\left(ml\right)\)
\(\Rightarrow m_{C_2H_5OH}=900.0,8=720\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{720}{46}=\dfrac{360}{23}\left(mol\right)\)
PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=\dfrac{360}{23}\left(mol\right)\)
Mà: H = 92%
\(\Rightarrow n_{CH_3COOH\left(TT\right)}=\dfrac{360}{23}.92\%=14,4\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=14,4.60=864\left(g\right)\)
b, \(m_{ddgiam}=\dfrac{864}{5\%}=17280\left(l\right)\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
1 0,5 ( mol )
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)
1 1 ( mol )
\(m_{CH_3COOH}=1.60.80\%=48g\)