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)

$n_{C_2H_5OH\ pư} = 0,4.80\% = 0,32(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$n_{CH_3COOH} = n_{C_2H_5OH\ pư} = 0,32(mol)$
$C_{M_{CH_3COOH}} = \dfrac{0,32}{0,2} = 1,6M$

\(m_{rượu} = 10.1000.0,8 = 8000(gam)\\ n_{rượu} = \dfrac{8000}{46} = \dfrac{4000}{23}(mol)\\ C_2H_5OH + O_2 \xrightarrow{t^o} CH_3COOH + H_2O\\ n_{CH_3COOH} = n_{C_2H_5OH} =\dfrac{4000}{23}.80\% = \dfrac{3200}{23}(mol)\\ C_{M_{CH_3COOH}} = \dfrac{\dfrac{3200}{23}}{10} =13,91M \)

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)

$V_{C_2H_5OH} = \dfrac{m}{D} = \dfrac{14,72}{0,8} = 18,4(ml)$

$V_{dd\ C_2H_5OH\ 20^o} = \dfrac{18,4.100}{20} = 92(ml)$

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: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)

\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)

\(n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right);n_{C_2H_5OH}=\dfrac{12}{46}=\dfrac{6}{23}\left(mol\right)\\ PTHH:CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ Vì:0,2:1< \dfrac{6}{23}:1\Rightarrow Ethanol.dư\\ n_{este\left(LT\right)}=n_{acid}=0,2\left(mol\right)\\ n_{este\left(TT\right)}=\dfrac{8}{88}=\dfrac{1}{11}\left(mol\right)\\ \Rightarrow H=\dfrac{\dfrac{1}{11}}{0,2}.100\%\approx45,455\%\)

\(CH_3COOH+C_2H_5OH⇌\left(H^+,t^o\right)CH_3COOC_2H_5+H_2O\\ n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right);n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\\ Vì:\dfrac{0,3}{1}>\dfrac{0,2}{1}\Rightarrow C_2H_5OH.dư\\ n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ n_{CH_3COOC_2H_5\left(TT\right)}=\dfrac{11}{88}=0,125\left(mol\right)\\ \Rightarrow H=\dfrac{0,125}{0,2}.100\%=62,5\%\)

\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25mol\)

\(C_6H_{12}O_6\underrightarrow{lênmen}2C_2H_5OH+2CO_2\)

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

\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{1}=0,5M\)

\({{\rm{n}}_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}}}{\rm{  =  }}\frac{{\rm{6}}}{{{\rm{60}}}}{\rm{  =  0,1 (mol); }}{{\rm{n}}_{{{\rm{C}}_2}{{\rm{H}}_5}{\rm{OH}}}}{\rm{  =  }}\frac{{{\rm{5,2}}}}{{46}}{\rm{ }} \approx {\rm{ 0,113 (mol)}}\)

Phương trình hóa học:

Ta có: \(\frac{{0,1}}{1} < \frac{{0,113}}{1}\) => acetic acid hết, ester tính theo acetic acid.

 \(\begin{array}{l}{{\rm{n}}_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COO}}{{\rm{C}}_2}{{\rm{H}}_5}}}{\rm{ =  }}{{\rm{n}}_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}}}{\rm{  =   0,1 (mol) }}\\ \Rightarrow {{\rm{m}}_{{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COO}}{{\rm{C}}_2}{{\rm{H}}_5}}} = {\rm{0,1}} \times {\rm{88  =  8,8 (g)}}\\ \Rightarrow {\rm{H  =  }}\frac{{5,28}}{{8,8}} \times 100\%  = 60\% \end{array}\)