a, \(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)

\(C_2H_5OH+K\rightarrow C_2H_5OK+\dfrac{1}{2}H_2\)

\(C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

b, \(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)

\(CH_3COOH+K\rightarrow CH_3COOK+\dfrac{1}{2}H_2\)

\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt: H₂SO₄ đặc, t^o)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

Giải giúp

- Tác dụng với CH₃COOH:

Mg + 2CH₃COOH ---> (CH₃COO)₂Mg + H₂

KOH + CH₃COOH ---> CH₃COOK + H₂O

2Na + 2CH₃COOH ---> 2CH₃COONa + H₂

K₂CO₃ + 2CH₃COOH ---> 2CH₃COOK + CO₂ + H₂O

- Tác dụng với C₂H₅OH:

2Na + 2C₂H₅OH ---> 2C₂H₅ONa + H₂

- Tác dụng với (RCOO)₃C₃H₅:

KOH + (RCOO)₃C₃H₅ ---> 3RCOOK + C₃H₅(OH)₃

\(n_{C_2H_4}=\dfrac{1,12}{22,4}=0,05mol\)

\(C_2H_4+H_2O\rightarrow\left(t^o,H_2SO_4\right)C_2H_5OH\)

0,05                                           0,05      ( mol )

\(C_2H_5OH+CH_3COOH\rightarrow\left(t^o,H_2SO_4\right)CH_3COOC_2H_5+H_2O\)

  0,05                                                               0,05                        ( mol )

\(m_{CH_3COOC_2H_5}=0,05.88=4,4g\)

$n_{CH_3COOH} = 0,75(mol)$
$n_{C_2H_5OH} = 1,5(mol)$

$n_{CH_3COOC_2H_5} = 0,4875(mol)$
$CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$

Ta thấy : 

$n_{CH_3COOH} : 1 < n_{C_2H_5OH} :1$ nên hiệu suất tính theo số mol $CH_3COOH$
$n_{CH_3COOH\ pư} = n_{CH_3COOC_2H_5} = 0,4875(mol)$
$\Rightarrow H = \dfrac{0,4875}{0,75}.100\% = 65\%$

Bài 1:

PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)

\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)

Bài 2:

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết

\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)