Bài 20:

- Sử dụng quỳ tím:

+ Rượu etylic không làm quỳ tím đổi màu.

+ Axit axetic làm quỳ tím chuyển đỏ.

- Sử dụng Na₂CO₃.

+ Có hiện tượng sủi bọt khí: axit axetat.

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

+ Không hiện tượng: rượu etylic

Bài 20:

\(C_2H_4+H_2O\underrightarrow{t^o,xt}C_2H_5OH\)

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

\(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)

\(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)

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

Câu 22:

\(n_{CH_3COOH}=\dfrac{45}{60}=0,75\left(mol\right)\)

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

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,75\left(mol\right)\)

Mà: H = 80%
\(\Rightarrow n_{CH_3COOC_2H_5\left(TT\right)}=0,75.80\%=0,6\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(TT\right)}=0,6.88=52,8\left(g\right)\)

Câu 23:

m_CaCO3 = 20.80% = 16 (g) \(\Rightarrow n_{CaCO_3}=\dfrac{16}{100}=0,16\left(mol\right)\)

PT: \(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\)

a, Theo PT: \(n_{CH_3COOH}=2n_{CaCO_3}=0,32\left(mol\right)\Rightarrow m_{CH_3COOH}=0,32.60=19,2\left(g\right)\)

b, \(n_{CO_2}=n_{CaCO_3}=0,16\left(mol\right)\Rightarrow V_{CO_2}=0,16.22,4=3,584\left(l\right)\)

a) $n_{(CH_3COO)_2Mg} = \dfrac{14,2}{142} = 0,1(mol)$

$2CH_3COOH + Mg \to (CH_3COO)_2Mg  + H_2$

Theo PTHH : $n_{CH_3COOH} = 2n_{(CH_3COO)_2Mg} = 0,2(mol)$
$C_{M_{CH_3COOH}} = \dfrac{0,2}{0,4} = 0,5M$

b) $n_{H_2} = n_{(CH_3COO)_2Mg} = 0,1(mol)$
$V_{H_2} = 0,1.22,4 = 2,24(lít)$

c)

$CH_3COOH + NaOH \to CH_3COONa + H_2O$
$n_{NaOH} = n_{CH_3COOH} = 0,1(mol)$
$V_{dd\ NaOH} = \dfrac{0,1}{0,75} = 0,13(lít)$

Bài 25:

\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

Theo PT: \(n_{CH_3COOH}=2n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{16,6}.100\%\approx72,29\%\\\%m_{C_2H_5OH}\approx27,71\%\end{matrix}\right.\)

Bài 26:

Ta có: \(46n_{C_2H_5OH}+60n_{CH_3COOH}=21,2\left(g\right)\left(1\right)\)

PT: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)

\(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\)

Theo PT: \(n_{H_2}=\dfrac{1}{2}n_{C_2H_5OH}+\dfrac{1}{2}n_{CH_3COOH}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\left(2\right)\)

Từ (1) và (2) \(\Rightarrow n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{0,2.46}{21,2}.100\%\approx43,4\%\\\%m_{CH_3COOH}\approx56,6\%\end{matrix}\right.\)

a) $C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
b) $C_2H_5OH + 3O_2 \xrightarrow{t^o} 2CO_2 + 3H_2O$
c) $C_2H_5OH + CH_3COOH \to CH_3COOC_2H_5 + H_2O$
d) $2C_2H_5OH + 2K \to 2C_2H_5OK + H_2O$
e) $CH_3COOH + KOH\to CH_3COOK + H_2O$
f) $CH_3COOC_2H_5 + H_2O \to CH_3COOH + C_2H_5OH$

g) $2CH_3COOH + Mg \to (CH_3COO)_2Mg + H_2O$

h) $2CH_3COOH + MgCO_3 \to (CH_3COO)_2Mg + CO_2 + H_2O$

$n_{CH_3COOH} = \dfrac{12}{60} = 0,2(mol)$
$n_{C_2H_5OH} = \dfrac{1,38}{46} = 0,03(mol)$
\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)

Ta thấy :

$n_{CH_3COOH} : 1 > n_{C_2H_5OH}:1$ nên hiệu suất tính theo số mol $C_2H_5OH$
$n_{CH_3COOC_2H_5} = n_{C_2H_5OH\ pư} = 0,03.75\% = 0,0375(mol)$

$m_{CH_3COOC_2H_5} = 0,0375.88 =3,3(gam)$

a)

1,5 lít = 1500 ml

$V_{rượu} = 1500.\dfrac{20}{100} = 300(ml)$

b)
$m_{rượu} = D.V = 0,8.300 = 240(gam)$

$n_{C_2H_5OH} = \dfrac{240}{46} = \dfrac{120}{23}(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

$n_{CH_3COOH} = n_{C_2H_5OH} = \dfrac{120}{23}(mol)$
$m_{CH_3COOH} = \dfrac{120}{23}.60 = 313,04(gam)$

$Na_2CO_3 + 2CH_3COOH \to 2CH_3COONa + CO_2 + H_2O$

$CO_2 + Ca(OH)_2 \to CaCO_3 + H_2O$

Theo PTHH : 

$n_{CO_2} = n_{CaCO_3} = \dfrac{10}{100} = 0,1(mol)$

$n_{CH_3COOH} = 2n_{CO_2} = 0,2(mol)$

$m_{CH_3COOH} = 0,2.60 = 12(gam)$

$2CH_3COOH + 2Na \to 2CH_3COONa + H_2$
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$

Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_{CH_3COOH} + \dfrac{1}{2}n_{C_2H_5OH} = 0,2(mol)$
$\Rightarrow n_{C_2H_5OH} = 0,2(mol)$

$m_{C_2H_5OH} = 0,2.46 = 9,2(gam)$

$\%m_{CH_3COOH} = \dfrac{12}{12 + 9,2}.100\% = 56,6\%$
$\%m_{C_2H_5OH} = 100\% - 56,6\% = 43,4\%$

cho C2H5OH và CH3COOH tác dụng với Na

+) làm Natri tan, sủi bọt khí là C2H5OH

+) chất còn lại là CH3COOH

2C2H5OH + 2Na ---> 2C2H5ONa + H2