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PTHH: \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\) (1)
\(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\) (2)
a) Ta có: \(n_{H_2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)
Đặt số mol của \(C_2H_5OH\) là \(a\) \(\Rightarrow n_{H_2\left(1\right)}=\frac{1}{2}a\)
Đặt số mol của \(CH_3COOH\) là \(b\) \(\Rightarrow n_{H_2\left(2\right)}=\frac{1}{2}b\)
Ta có hệ phương trình:
\(\left\{{}\begin{matrix}46a+60b=20,5\\\frac{1}{2}a+\frac{1}{2}b=0,2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,25\\b=0,15\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}n_{C_2H_5OH}=0,25mol\\n_{CH_3COOH}=0,15mol\end{matrix}\right.\)
\(\Rightarrow m_{CH_3COOH}=60\cdot0,15=9\left(g\right)\)
\(\Rightarrow\%m_{CH_3COOH}=\frac{9}{20,5}\cdot100\approx43,9\%\)
\(\Rightarrow\%m_{C_2H_5OH}=56,1\%\)
b) PTHH: \(C_2H_5OH+CH_3COOH\underrightarrow{xt}CH_3COOC_2H_5+H_2O\)
Xét tỷ lệ: \(\frac{0,15}{1}< \frac{0,25}{1}\) \(\Rightarrow\) Axit phản ứng hết, Rượu còn dư
\(\Rightarrow n_{CH_3COOC_2H_5}=0,15mol\) \(\Rightarrow m_{este}=0,15\cdot88=13,2\left(g\right)\)
\(\Rightarrow m_{este}thực=13,2\cdot90\%=11,88\left(g\right)\)
\(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
\(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\)
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
=> \(\left\{{}\begin{matrix}\%_{m_{CH_3COOH}}=\dfrac{60.0,1.100}{10,6}=56,6\%\\\%_{m_{C_2H_5OH}}=100-56,6=43,4\%\end{matrix}\right.\)
\(m_{muối}=m_{CH_3COONa}+m_{C_2H_5ONa}=82.0,1+68.0,1=15\left(g\right)\)
100 - 56,6 sao bằng 43,4%
Xem lại đơn vị
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) 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 --H2SO4(đ),to--> CH3COOC2H5 + H2O
Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH3COOH
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,15<---------------------------------0,15
=> \(H=\dfrac{0,15}{0,2}.100\%=75\%\)
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
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 \(\left\{{}\begin{matrix}CH_3COONa:0,2\left(mol\right)\\C_2H_5OH:\dfrac{39,6-12}{46}=0,6\left(mol\right)\end{matrix}\right.\)
\(V_{dd}=\dfrac{0,1}{2}=0,05\left(l\right)\)
=> \(\left\{{}\begin{matrix}C_{M\left(CH_3COONa\right)}=\dfrac{0,2}{0,05}=4M\\C_{M\left(C_2H_5OH\right)}=\dfrac{0,6}{0,05}=12M\end{matrix}\right.\)
chết làm thiếu nha bạn
để mình sửa lại nha:D