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Câu 1:
Gọi \(\left\{{}\begin{matrix}n_{CH3COOH}:a\left(mol\right)\\n_{C2H5OH}:b\left(mol\right)\end{matrix}\right.\)
PTHH:
\(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
Giải hệ PT:
\(\left\{{}\begin{matrix}60a+46b=27,2\\a+b=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,3\\b=0,2\end{matrix}\right.\)
\(\Rightarrow n_{CH3COOC2H5}=n_{C2H5OH}=0,2\left(mol\right)\)
\(\Rightarrow m_{C2H5COOC2H5}=0,2.88=17,6\left(g\right)\)
Câu 2:
a, PTHH:
\(2Na+2CH_3COOH\rightarrow2CH_3COONa+H_2\)
\(2Na+2C_2H_5OH\rightarrow2C_2H_5ONa+H_2\)
\(CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)
b, Ta có:
\(n_{CO2}=\frac{2,24}{22,4}=0,1\left(mol\right)\)
\(n_{H2}=\frac{5,6}{22,4}=0,25\left(mol\right)\)
\(\Rightarrow n_{CH3COOH}=2n_{CO2}=0,1.2=0,2\left(mol\right)\)
\(\Rightarrow n_{Hh}=0,25.2=0,5\left(mol\right)\Rightarrow n_{C2H5OH}=0,5-0,2=0,3\left(mol\right)\)
\(\Rightarrow m=25,8\left(g\right)\)
c,\(PTHH:C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)
\(\Rightarrow n_{CH3COOC2H5}=n_{CH3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{CH3COOC2H5}=14,08\left(g\right)\)
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
=> mCH3COOH = 0,5.60 = 30 (g)
=> mC2H5OH = 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
=> mNaHCO3 = 0,25.84 = 21 (g)
- 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. C2H5OH + Zn ---/----> Ko xảy ra
2CH3COOH + Zn --------> (CH3COOH)2Zn + H2
b. Ta có :
n hỗn hợp = 6,72/22,4=0,3 mol
Theo PTHH: n CH3COOH = 2 n H2 =0,6 mol
=> m CH3COOH = 0,6.60=36g
=> m C2H5OH =60-36=24g
=> % m C2H5OH= 24/60.100%=40%
=>% m CH3COOH = 100%-40%=60%
c. Ta có:
V rượu = 24/0,8=30 ml
=>V dd= 30.100%/15*=200ml
nH2 = 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%
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%
CH3COOH +Na =>CH3COONa +1/2H2
x mol =>0,5x mol
C2H5OH+ Na=>C2H5ONa +1/2H2
y mol. =>0,5y mol
nH2=5,6/22,4=0,25 mol=>x+y=0,5
mhh bđ=60x+46y=27,2
=>x=0,3 và y=0,2
mCH3COOH=0,3.60=18g=>%mCH3COOH=18/27,2.100%=66,17%
%mC2H5OH=33,82%