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

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

    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:>

\(Đặt:n_{C_2H_5OH}=a\left(mol\right);n_{CH_3COOH}=b\left(mol\right)\left(a,b>0\right)\\ a,PTHH:C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\\ CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\\ n_{H_2\left(tổng\right)}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ b,Ta.lập.hpt:\left\{{}\begin{matrix}46a+60b=10,6\\0,5a+0,5b=0,1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,1\\b=0,1\end{matrix}\right.\\ \%m_{C_2H_5OH}=\dfrac{0,1.46}{10,6}.100\approx43,396\%\\\Rightarrow\%m_{CH_3COOH}\approx100\%-43,396\%\approx56,604\%\)

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

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

 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

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

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

Rượu etylic \(C_2H_5OH\)

Axit axetic \(CH_3COOH\)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1mol\)

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

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

\(\%m_{CH_3COOH}=\dfrac{0,2\cdot60}{39,6}\cdot100\%=30,3\%\)      

\(\%m_{C_2H_5OH}=100\%-30,3\%=69,7\%\)

a)

\(n_{CO_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

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

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

=> \(m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\%m_{CH_3COOH}=\dfrac{12}{39,6}.100\%=30,3\%\)

\(\%m_{C_2H_5OH}=\dfrac{39,6-12}{39,6}.100\%=69,7\%\)

b) dd sau pư chứa \(\left\{{}\begin{matrix}CH_3COONa:0,2\left(mol\right)\\C_2H_5OH:\dfrac{39,6-12}{46}=0,6\left(mol\right)\end{matrix}\right.\)

\(V_{dd}=\dfrac{0,1}{2}=0,05\left(l\right)\)

=> \(\left\{{}\begin{matrix}C_{M\left(CH_3COONa\right)}=\dfrac{0,2}{0,05}=4M\\C_{M\left(C_2H_5OH\right)}=\dfrac{0,6}{0,05}=12M\end{matrix}\right.\)

a, Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{CH_3COOH}=b\left(mol\right)\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

PTHH: 

2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂

a---------------------------------------->0,5a

2CH₃COOH + 2Na ---> 2CH₃COONa + H₂

b------------------------------------------------>0,5b

=> hệ pt \(\left\{{}\begin{matrix}46a+60b=48,8\\0,5a+0,5b=0,5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,8\left(mol\right)\\b=0,2\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{C_2H_5OH}=0,8.46=36,8\left(g\right)\\m_{CH_3COOH}=0,2.60=12\left(g\right)\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{36,8}{48,8}.100\%=75,41\%\\\%m_{CH_3COOH}=100\%-75,41\%=24,59\%\end{matrix}\right.\)

b, PTHH:

\(C_2H_5OH+CH_3COOH\xrightarrow[t^o]{H_2SO_4đặc}CH_3COOC_2H_5+H_2O\)

LTL: 0,8 > 0,2 => Rượu dư

\(n_{CH_3COOC_2H_5\left(tt\right)}=0,2.85\%=0,17\left(mol\right)\\ m_{este}=0,17.88=14,96\left(g\right)\)

a.Gọi \(\left\{{}\begin{matrix}n_{C_2H_5OH}=x\\n_{CH_3COOH}=y\end{matrix}\right.\)

\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)

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

             x                                          1/2 x         ( mol )

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

        y                                                      1/2 y       ( mol )

Ta có:

\(\left\{{}\begin{matrix}46x+60y=48,8\\\dfrac{1}{2}x+\dfrac{1}{2}y=0,5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=0,8\\y=0,2\end{matrix}\right.\)

\(\rightarrow m_{C_2H_5OH}=0,8.46=36,8g\)

\(\rightarrow\left\{{}\begin{matrix}\%m_{C_2H_5OH}=\dfrac{36,8}{48,8}.100=75,4\%\\\%m_{CH_3COOH}=100\%-75,4\%=24,6\%\end{matrix}\right.\)

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

        0,8     <          0,2                                                                       ( mol )

                               0,2                                             0,2                    ( mol )

\(m_{CH_3COOC_2H_5}=0,2.88.85\%=14,96g\) 

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)

2C2H5OH + 2Na--> 2C2H5Na + H2

  a                                                 a/2           mol

2CH3COOH + 2Na --> 2CH3COONa + H2

 b                                                             b/2             mol

n khí = 3,36/22,4=0,15 mol

=> a/2 + b/2 =0,15

và 46a + 60 b =15,2 

=> a=0,2 mol  : b=0,1 mol

=> mC2H5OH = 0,2 * 46=9,2 g

=>% mC2H5OH = 9,2*100/15,2=60,53%

% mCH3COOH = 100 - 60 ,53=39,47 %