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\(n_{NaOH}=0,2.1=0,2\left(mol\right)\\ a,CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\\ n_{CH_3COOH}=n_{NaOH}=0,2\left(mol\right)\\ b,m_{CH_3COOH}=0,2.60=12\left(g\right)\\ m_{C_2H_5OH}=20-12=8\left(g\right)\)
- 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)
2C2H5OH+ 2Na=× 2C2H5ONa + H2
2CH3COOH + 2Na=× 2CH3COONa + H2
b)
C2H4 + Br2=× C2H4Br2
Fe, t
C6H6 + Br2 = × C6H5Br+ HBr
c)
2CH3COOH+ Na2CO3=× 2CH3COONa+H2O+CO2
nNaOH= 0.1*0.2=0.02 mol
CH3COOH + NaOH --> CH3COONa + H2O
=> nCH3COOH= 0.02 mol
mCH3COOH= 1.2g
nH2= 0.336/22.4=0.015 mol
CH3COOH + Na --> CH3COONa + 1/2H2 (1)
C2H5OH + Na --> C2H5ONa + 1/2 H2
nH2(1)= 0.01 mol
nH2(2)= 0.015-0.01=0.005 mol
=> nC2H5OH= 0.01 mol
mC2H5OH= 0.46g
mX= mCH3COOH + mC2H5OH= 1.12+0.46=1.58g
a) nNaOH = 0,6.1 = 0,6 (mol)
PTHH: NaOH + CH3COOH --> CH3COONa + H2O
0,6----->0,6
=> mCH3COOH = 0,6.60 = 36 (g)
=> mC2H5OH = 45,2 - 36 = 9,2 (g)
b) \(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)
PTHH: 2CH3COOH + 2Na --> 2CH3COONa + H2
0,6---------------------------->0,3
2C2H5OH + 2Na --> 2C2H5ONa + H2
0,2--------------------------->0,1
=> V = (0,3 + 0,1).22,4 = 8,96 (l)
a) V(rượu)= 200.35/100= 70(ml)
=> V(H2O)= 200-70=130(ml)
b) Na + H2O -> NaOH +1/2 H2
C2H5OH + Na -> C2H5ONa + 1/2 H2
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