cho hhM gồm axit axetic và rượu etylic td hết với natri dư thu được 0,672 lít khí(đktc).Nếu cho hh trên td với dd NaOH thì cần 0,04mol NaOH.Tính khối lượng rượu etylic ban đầu
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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 %
\(a,n_{H_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\\ n_{NaOH}=0,2.0,1=0,02\left(mol\right)\)
PTHH:
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
0,02<-----------0,02
\(CH_3COOH+Na\rightarrow CH_3COOH+\dfrac{1}{2}H_2\uparrow\)
0,02------------------------------------------>0,01
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)
0,07<-----------------------------------0,14
\(\rightarrow m=0,01.60+0,07.46=3,82\left(g\right)\)
\(b,\left\{{}\begin{matrix}\%m_{CH_3COOH}=\dfrac{0,01.60}{3,82}.100\%=15,71\%\\\%m_{C_2H_5OH}=100\%-15,71\%=84,29\%\end{matrix}\right.\)
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)
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 )
\(m_{CH_3COOC_2H_5}=0,2.88.85\%=14,96g\)
\(n_{H_2}=\dfrac{11,2}{22,4}=0,5mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
1 0,5 ( mol )
\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)
1 1 ( mol )
\(m_{CH_3COOH}=1.60.80\%=48g\)
n CH3COOH = a(mol) ; n C2H5OH = b(mol)
=> 60a + 46b = 21,2(1)
$2CH_3COOH + 2Na \to 2CH_3COONa + H_2$
$2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2$
Theo PTHH :
n H2 = 0,5a + 0,5b = 4,48/22,4 = 0,2(2)
Từ (1)(2) suy ra a = b = 0,2
m CH3COOH = 0,2.60 = 12(gam)
m C2H5OH = 0,2.46 = 9,2(gam)
\(n_{H_2}=\dfrac{0,672}{22,4}=0,03\left(mol\right)\)
PTHH:
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
0,04<------------0,04
\(CH_3COOH+Na\rightarrow CH_3COONa+\dfrac{1}{2}H_2\uparrow\)
0,04----------------------------------------->0,02
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\uparrow\)
0,005<--------------------------------0,01
\(\rightarrow m_{C_2H_5OH}=0,005.46=0,23\left(g\right)\)