Bài 45. Axit axetic

\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)

PTHH: MgO + 2CH₃COOH ---> (CH₃COO)₂Mg + H₂O

          0,2-------->0,4------------------>0,2

=> \(\left\{{}\begin{matrix}m_{ddCH_3COOH}=\dfrac{0,4.60}{10\%}=240\left(g\right)\\m_{muối}=0,2.142=28,4\left(g\right)\end{matrix}\right.\)

Các chất tác dụng với axit axetic \(CH_3COOH\):

\(Na_2O+2CH_3COOH\rightarrow2CH_3COONa+H_2O\)

\(C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)

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

\(2CH_3COOH+Cu\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Cu+2H_2O\)

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

a) \(n_{CH_3COOH}=\dfrac{100.12\%}{60}=0,2\left(mol\right)\)

PTHH: CH₃COOH + NaHCO₃ --> CH₃COONa + CO₂ + H₂O

                 0,2-------->0,2----------->0,2--------->0,2

=> m_NaHCO3 = 0,2.84 = 16,8 (g)

=> \(m_{dd.NaHCO_3}=\dfrac{16,8.100}{8,4}=200\left(g\right)\)

b) m_CH3COONa = 0,2.82 = 16,4 (g)

m_{dd sau pư} = 100 + 200 - 0,2.44 = 291,2 (g)

=> \(C\%_{dd.muối}=\dfrac{16,4}{291,2}.100\%=5,632\%\)

6CH3COOH+Fe2O3->2(CH3COO)3Fe+3H2O

=>Fe2O3 tan , dd có màu nâu đỏ 

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\(PTHH:6C_2H_4O_2+Fe_2O_3\rightarrow2Fe\left(CH_3COO\right)_3+3H_2O\)

\(HT:Fe_2O_3\) tan, dd chuyển màu nâu đỏ

\(m_{CH_3COOH}=6\%.200=12\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{12}{60}=0,2\left(mol\right)\\ n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

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

LTL: \(\dfrac{0,2}{2}>0,2\rightarrow\) Zn dư

Theo pthh: \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn\left(pư\right)}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=\dfrac{1}{2}.0,2=0,1\left(mol\right)\)

\(\rightarrow\left\{{}\begin{matrix}m_{H_2}=0,1.2=0,2\left(g\right)\\m_{Zn\left(pư\right)}=0,1.65=6,5\left(g\right)\\m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\end{matrix}\right.\)

\(\rightarrow m_{dd}=200+6,5-0,2=206,3\left(g\right)\\ \rightarrow C\%_{\left(CH_3COO\right)_2Zn}=\dfrac{18,3}{206,3}=8,87\%\)

\(n_{K2CO3}=\dfrac{13,8}{138}=0,1\left(mol\right)\)

a) Pt : \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O|\)

                     2                  1                     2                1            1

                    0,2              0,1                                     0,1

b) \(n_{CO2}=\dfrac{0,1.1}{1}=0,1\left(mol\right)\)

\(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

c) \(n_{CH3COOH}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

\(V_{ddCH3COOH}=\dfrac{0,2}{2}=0,1\left(l\right)\)

 Chúc bạn học tốt

a.b.\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25mol\)

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

       0,5                                                              0,25               ( mol )

\(m_{CH_3COOH}=0,5.60=30g\)

\(\%m_{CH_3COOH}=\dfrac{30}{45}.100=66,67\%\)

\(\%m_{C_2H_5OH}=100\%-66,67\%=33,33\%\)

c.\(m_{NaOH}=50.20\%=10g\)

\(n_{NaOH}=\dfrac{10}{40}=0,25mol\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

 \(\dfrac{0,25}{2}\)   <    \(\dfrac{0,25}{1}\)                                ( mol )

 0,25                           0,125                    ( mol )

\(m_{Na_2CO_3}=0,125.106=13,25g\)

a) 

2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ --> 2CH₃COONa + CO₂ + H₂O

                  0,5<-----------------------------------0,25

=> m_CH3COOH = 0,5.60 = 30 (g)

=> m_C2H5OH = 45 - 30 = 15 (g)

c) \(n_{NaOH}=\dfrac{50.20\%}{40}=0,25\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{0,25}{0,25}=1\) => Tạo muối NaHCO₃

PTHH: NaOH + CO₂ --> NaHCO₃

              0,25-------------->0,25

=> m_NaHCO3 = 0,25.84 = 21 (g)

đây có phải chương trinhg mới ko hay cũ

2CH3COOH+Fe->(CH3COO)2Fe+H2

0,4------------------0,2----0,2--------------0,2

n H2=\(\dfrac{4,958}{24,79}\)=0,2 mol

=>CM CH3COOH=\(\dfrac{0,4}{0,2}\)=2M

=>m (CH3COO)2Fe=0,2.174=34,8g

CH3COOH+NaOH->CH3COONa+H2O

0,1---------------0,1

n CH3COOH=2.0,05=0,1 mol

=>m NaOH=0,1.40=4g

=>m dd NAOH=400g