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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 C2H5OH =a (mol) ; n CH3COOH = b(mol)
=> 46a + 60b = 27,2(1)
$2C_2H_5ONa + 2Na \to 2C_2H_5ONa + H_2$
$2CH_3COOH + 2Na \to 2CH_3COONa + H_2$
Theo PTHH :
n H2 = 0,5a + 0,5b = 5,6/22,4 = 0,25(2)
Từ (1)(2) suy ra a = 0,2 ; b = 0,3
Suy ra:
m C2H5OH = 0,2.46 = 9,2(gam)
m CH3COOH = 0,3.60 = 18(gam)
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 %
- Đặt \(\left\{{}\begin{matrix}n_{C_2H_5OH}=a\left(mol\right)\\n_{CH_3COOH}=b\left(mol\right)\end{matrix}\right.\) \(\Rightarrow46a+60b=33,2\left(1\right)\)
\(n_{H_2}=\dfrac{V}{22,4}=\dfrac{6,72}{22,4}=0,3\left(mol\right)\)
a) \(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
2 2 1 (mol)
a a a/2 (mol)
\(2CH_3COOH+2Na\rightarrow2CH_3COONa+H_2\)
2 2 1 (mol)
b b b/2 (mol)
Từ hai PTHH trên ta có: \(\dfrac{a}{2}+\dfrac{b}{2}=n_{H_2}=0,3\Rightarrow a+b=0,6\left(2\right)\)
(1), (2) ta có hệ phương trình: \(\left\{{}\begin{matrix}46a+60b=33,2\\a+b=0,6\end{matrix}\right.\)
Giải ra ta được: \(a=0,2\left(mol\right);b=0,4\left(mol\right)\)
b) \(m_{C_2H_5OH}=n.M=0,2\times46=9,2\left(g\right)\)
\(m_{CH_3COOH}=n.M=0,4\times60=24\left(g\right)\)
c) \(m_{C_2H_5ONa}=n.M=0,2\times68=13,6\left(g\right)\)
\(m_{CH_3COONa}=n.M=0,4\times82=32,8\left(g\right)\)
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
=> \(H=\dfrac{0,15}{0,2}.100\%=75\%\)
Gọi \(\left\{{}\begin{matrix}n_{ancol}:x\left(mol\right)\\n_{axit}:y\left(mol\right)\end{matrix}\right.\)
PTHH:
\(C_2H_5OH+K\rightarrow C_2H_5ONa+\frac{1}{2}H_2\)
x_____________________________0,5x
\(CH_3COOH+K\rightarrow CH_3COONa+\frac{1}{2}H_2\)
y____________________________________0,5y
Giải hệ PT:
\(\left\{{}\begin{matrix}46x+60y=20,5\\0,5x+0,5y=\frac{4,48}{22,4}=0,2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0,25\\y=0,15\end{matrix}\right.\)
\(PTHH:C_2H_5OH+CH_3COOH⇌CH_3COOC_2H_5+H_2O\)
\(\Rightarrow n_{CH3COOC2H5}=n_{CH3COOH}=0,15\left(mol\right)\)
\(\Rightarrow m_{este}=13,2\left(g\right)\)