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

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

 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

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

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

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

a) Gọi số mol CH₃COOH, C₂H₅OH là a, b (mol)

=> 60a + 46b = 25,8 (1)

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

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

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

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

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

=> 0,5a + 0,5b = 0,25 (2)

(1)(2) => a = 0,2 (mol); b = 0,3 (mol)

=> \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,2.60}{25,8}.100\%=46,51\%\\\%m_{C_2H_5OH}=\dfrac{0,3.46}{25,8}.100\%=53,49\%\end{matrix}\right.\)

b) 

\(n_{CH_3COOC_2H_5}=\dfrac{13,2}{88}=0,15\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đ),t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

PTHH: CH₃COOH + C₂H₅OH --H₂SO_4(đ),t^o--> CH₃COOC₂H₅ + H₂O

                   0,15<---------------------------------0,15

=> \(H=\dfrac{0,15}{0,2}.100\%=75\%\)

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 )

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

\(a,n_{NaOH}=1,5.0,2=0,3\left(mol\right)\\ n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

PTHH: 

CH₃COOH + NaOH ---> CH₃COONa + H₂O

0,3<-----------0,3

2CH₃COOH + 2Na ---> 2CH₃COONa + H₂

0,3----------------------------------------------->0,15

2C₂H₅OH + 2Na ---> 2C₂H₅ONa + H₂

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

=> m = 0,2.46  +0,3.60 = 27,2 (g)

b) \(\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,3.60}{27,2}.100\%=66,18\%\\\%m_{C_2H_5OH}=100\%-66,18\%=33,82\%\end{matrix}\right.\)