\(C_2H_5OH+K_2CO_3\rightarrow\left(kopứ\right)\)

\(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)

    2                      1                  2                    1           1       (mol)

  0,4                     0,2              0,4                0,2         0,2 (mol)

\(nCO_2=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

\(mCH_3COOH=0,4.60=24\left(g\right)\)

\(mK_2CO_3=0,2.138=27,6\left(g\right)\)

\(mCH_3COOK=0,4.98=39,2\left(g\right)\)

\(mCO_2=0,2.44=8,8\left(g\right)\)

\(mdd=mCH_3COOH+mK_2CO_3+mCH_3COOK-mCO_2\)

\(=24+27,6+39,2-8,8=82\left(g\right)\)

\(C\%m_{CH_3COOH}=\dfrac{24.100}{82}=29,27\%\)

\(C\%m_{K_2CO_3}=\dfrac{27,6.100}{82}=33,66\%\)

câu thứ 2 bn tự lm cho bt:>

c2h5oh+na2co3-

2ch3cooh+ na2co3-> 2ch3coona+h2o+co2

nCO2=2,24/22,4=0,1

nCh3cooh=2nCO2 =2*0,1=0,2mol

mCH3COOH=0,2*60=12g

mc2h5oh=30,4-12=18,4

%mCH3COOH=12/30,4*100=39,5%

%mC2H5OH=18,4/30,4*100=60,5%

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

nCO2= 2.24/22.4=0.1 (mol)

Từ PTHH :

nCH3COOH= 0.2 (mol)

mCH3COOH= 0.2*60=12g

mC2H5OH= 30.4-12=18.4g

%mCH3COOH= 12/30.48*100%=39.47%

%mC2H5OH= 100-39.47=60.53%

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

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

 0,1                                0,1

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

0,1                           0,1

Theo pthh có: \(n_A=2nH_2=2.0,1=0,2\left(mol\right)\)

Gọi x, y là số mol của rượu và axit có trong hh A.

có hệ: \(\left\{{}\begin{matrix}x+y=0,2\\60x+46y=10,6\end{matrix}\right.\)

=> x = y = 0,1

=> \(\left\{{}\begin{matrix}\%_{m_{CH_3COOH}}=\dfrac{60.0,1.100}{10,6}=56,6\%\\\%_{m_{C_2H_5OH}}=100-56,6=43,4\%\end{matrix}\right.\)

\(m_{muối}=m_{CH_3COONa}+m_{C_2H_5ONa}=82.0,1+68.0,1=15\left(g\right)\)

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%

PT; C₂H₅0H + Na \(\rightarrow C_2H_5ONa\) + \(\dfrac{1}{2}\) H₂

CH₃COOH + Na \(\rightarrow\) CH₃COONa + \(\dfrac{1}{2}\)H₂

CH3COOH + NaHCO3 => CH3COONa + CO2 + H2O

nH2 = V/22.4 = 4.48/22.4 = 0.2 (mol)

Theo phương trình suy ra: nCH3COOH = 0.2 (mol)

mCH3COOH = n.M = 0.2 x 60 = 12 (g)

mC2H5OH = 30.4 - 12 = 18.4 (g)

%CH3COOH = 39.47%, %C2H5OH = 60.57%

nCO2= 0.2 mol

CH3COOH + NaHCO3 --> CH3COONa + CO2 + H2O

0.2_________________________________0.2

mCH3COOH= 12g

mC2H5OH= 18.4g

%CH3COOH= 39.47%

%C2H5OH= 60.53%

CH 3 COOH + NaOH →  CH 3 COONa +  H 2 O

C n H 2 n + 1 COOH + NaOH →  C n H 2 n + 1 COONa +  H 2 O