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)

Axit axetic \(CH_3COOH\)

Rượu etylic \(C_2H_5OH\)

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

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

0,2                                                               0,1

\(m_{CH_3COOH}=0,2\cdot60=12g\)

\(\%m_{CH_3COOH}=\dfrac{12}{20}\cdot100\%=60\%\)

\(\%m_{C_2H_5OH}=100\%-60\%=40\%\)

Câu 1:

Gọi \(\left\{{}\begin{matrix}n_{CH3COOH}:a\left(mol\right)\\n_{C2H5OH}:b\left(mol\right)\end{matrix}\right.\)

PTHH:

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

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

Giải hệ PT:

\(\left\{{}\begin{matrix}60a+46b=27,2\\a+b=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)

\(\Rightarrow n_{CH3COOC2H5}=n_{C2H5OH}=0,2\left(mol\right)\)

\(\Rightarrow m_{C2H5COOC2H5}=0,2.88=17,6\left(g\right)\)

Câu 2:

a, PTHH:

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

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

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

b, Ta có:

\(n_{CO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)

\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow n_{CH3COOH}=2n_{CO2}=0,1.2=0,2\left(mol\right)\)

\(\Rightarrow n_{Hh}=0,25.2=0,5\left(mol\right)\Rightarrow n_{C2H5OH}=0,5-0,2=0,3\left(mol\right)\)

\(\Rightarrow m=25,8\left(g\right)\)

c,\(PTHH:C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)

\(\Rightarrow n_{CH3COOC2H5}=n_{CH3COOH}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH3COOC2H5}=14,08\left(g\right)\)

a) 

C2H5OH + Na → C2H5ONa + 1/2 H2

CH3COOH + Na → CH3COONa + 1/2 H2

b)

Theo PTHH :

n C2H5OH = a(mol) ; n CH3COOH = b(mol)

=> 46a + 60b = 12,9(1)

n H2 = 0,5a + 0,5b = 2,8/22,5 = 0,125(2)

Từ (1)(2) suy ra a = 0,15 ; b = 0,1

%m C2H5OH = 0,15.46/12,9  .100% = 53,49%

%m CH3COOH = 100% -53,49% = 46,51%

b)

n C2H5ONa = a = 0,15 mol

n CH3COONa = b = 0,1(mol)

=> m muối = 0,15.69 + 0,1.82 = 18,55 gam

n_H2 = 0,2 mol

Gọi x, y lần lượt là số mol của C₂H₅OH, CH₃COOH ( x,y > 0 )

2C₂H₅OH + 2Na \(\rightarrow\) 2C₂H₅ONa + H₂

x...................x................x................0,5x

2CH₃COOH + 2Na \(\rightarrow\) 2CH₃COONa + H₂

y.........................y.................y................0,5y

Ta có hệ

\(\left\{{}\begin{matrix}46x+60y=21,2\\0,5x+0,5y=0,2\end{matrix}\right.\)

\(\Rightarrow\) \(\left\{{}\begin{matrix}x=0,2\\y=0,2\end{matrix}\right.\)

\(\Rightarrow\) %_C2H5OH = \(\dfrac{0,2.46.100\%}{21,2}\)\(\approx\) 43,4%

\(\Rightarrow\) %_CH3COOH = \(\dfrac{0,2.60.100\%}{21,2}\) \(\approx\) 56,6%

a. C2H5OH + Zn ---/----> Ko xảy ra

2CH3COOH + Zn --------> (CH3COOH)2Zn + H2

b. Ta có :

n hỗn hợp = 6,72/22,4=0,3 mol

Theo PTHH: n CH3COOH = 2 n H2 =0,6 mol

=> m CH3COOH = 0,6.60=36g

=> m C2H5OH =60-36=24g

=> % m C2H5OH= 24/60.100%=40%

=>% m CH3COOH = 100%-40%=60%

c. Ta có:

V rượu = 24/0,8=30 ml

=>V dd= 30.100%/15*=200ml

2Na+2H2O->2NaOH+H2

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

Na2O+H2O->2NaOH

0,1--------0,1-----------0,2

n H2=0,25 mol

=>m Na =0,5.23=11,5g

=>m Na2O=6,2g=>n Na2O=0,1 mol

=>m NaOH=0,7.40=28g

=>VH2O=0,6.22,4=13,44l

giúp mình với

\(n_{H_2}=\dfrac{5,6}{22,4}=0,25mol\)

\(2Na+2H_2O\rightarrow2NaOH+H_2\)

0,5                                        0,25

a)\(m_{Na}=\left(\right)\cdot23=11,5g\)

\(m_{Na_2O}=17,7-11,5=6,2g\)

b)\(n_{Na_2O}=\dfrac{6,2}{62}=0,1mol\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,1                           0,2

\(\Rightarrow\Sigma n_{bazo}=0,5+0,2=0,7mol\Rightarrow m_{NaOH}=0,7\cdot40=28g\)

c)\(m_{H_2O}=\left(0,5+0,1\right)\cdot18=10,8g\)