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a) $n_{C_6H_{12}O_6} = \dfrac{36}{180} = 0,2(mol)$
$n_{glucose\ pư} = 0,2.80\% = 0,16(mol)$
$C_6H_{12}O_6 \xrightarrow{t^o,men\ rượu} 2CO_2 + 2C_2H_5OH$
$n_{C_2H_5OH} = 2n_{glucose} = 0,32(mol)$
$m_{C_2H_5OH} = 0,32.46 = 14,72(gam)$
b)
$V_{C_2H_5OH} = \dfrac{m}{D} = \dfrac{14,72}{0,8} = 18,4(ml)$
$V_{dd\ C_2H_5OH\ 20^o} = \dfrac{18,4.100}{20} = 92(ml)$
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ V_{C_2H_5OH}=10000.8:100=800\left(ml\right)\\ m_{C_2H_5OH}=0,8.800=640\left(g\right)\\ m_{CH_3COOH}=\dfrac{60}{46}.640.80\%=\dfrac{30720}{46}\left(g\right)\\ m_{10lethanol}=640+9200.1=9840\left(g\right)\\ m_{O_2}=\dfrac{640.32}{46}=\dfrac{20480}{46}\left(g\right)\\ C\%_{ddCH_3COOH}=\dfrac{\dfrac{30720}{46}}{9840+\dfrac{20480}{46}}.100\%\approx6,493\%\)
a, \(n_{K_2CO_3}=\dfrac{2,76}{138}=0,02\left(mol\right)\)
PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)
Theo PT: \(n_{CH_3COOH}=2n_{K_2CO_3}=0,04\left(mol\right)\)
\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,04.60}{50}.100\%=4,8\%\)
b, \(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)
\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)
\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)
\(a,n_{C_6H_{12}O_6}=\dfrac{450}{180}=2,5\left(mol\right)\)
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: \(n_{CH_3COOH}=5\cdot80\%=4mol\)
\(m_{CH_3COOH}=4\cdot60=240g\)
\(m_{rượu} = 10.1000.0,8 = 8000(gam)\\ n_{rượu} = \dfrac{8000}{46} = \dfrac{4000}{23}(mol)\\ C_2H_5OH + O_2 \xrightarrow{t^o} CH_3COOH + H_2O\\ n_{CH_3COOH} = n_{C_2H_5OH} =\dfrac{4000}{23}.80\% = \dfrac{3200}{23}(mol)\\ C_{M_{CH_3COOH}} = \dfrac{\dfrac{3200}{23}}{10} =13,91M \)
\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25mol\)
\(C_6H_{12}O_6\underrightarrow{lênmen}2C_2H_5OH+2CO_2\)
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
\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{1}=0,5M\)
a)
$V_{C_2H_5OH} = 200.\dfrac{11,5}{100} = 23(ml)$
$m_{C_2H_5OH} = D.V = 0,8.23 = 18,4(gam)$
$n_{C_2H_5OH} = \dfrac{18,4}{46} = 0,4(mol)$
b)
$n_{C_2H_5OH\ pư} = 0,4.80\% = 0,32(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$n_{CH_3COOH} = n_{C_2H_5OH\ pư} = 0,32(mol)$
$C_{M_{CH_3COOH}} = \dfrac{0,32}{0,2} = 1,6M$