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\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1mol\)
\(2C_2H_5OH+2Na\rightarrow2C_2H_5ONa+H_2\)
0,1 0,1 0,05
\(V_{H_2}=0,05\cdot22,4=1,12l\)
\(m_{C_2H_5ONa}=0,1\cdot68=6,8g\)
Đáp án: B
n C 2 H 5 O H = 23 46 = 0 , 5 m o l
2 C 2 H 5 O H + 2 N a → 2 C 2 H 5 O N a + H 2 ↑
0,5 mol → 0,25 mol
⇒ V H 2 = 0 , 25 . 22 , 4 = 5 , 6
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) 2Na + 2C2H5OH \(\rightarrow\) 2C2H5ONa + H2
b) nNa = 0,23 : 23 = 0,01 mol
Theo pt: nH2 = \(\dfrac{1}{2}nNa=0,005mol\)
=> V H2 = 0,005.22,4 = 0,0112 lít
\(n_{CH_3COOH}=\dfrac{300\cdot5\%}{60}=0.25\left(mol\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(0.25........................................................0.125\)
\(V_{H_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(n_{C_2H_5OH}=0.1\cdot2=0.2\left(mol\right)\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(ĐK:H_2SO_{4\left(đ\right)},t^0\right)\)
\(0.2......................0.2.....................0.2\)
\(\Rightarrow CH_3COOHdư\)
\(m_{CH_3COOC_2H_5}=0.2\cdot88=17.6\left(g\right)\)
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 %
Bài 1:
nCH3COOH = 0,08.1,5 = 0,12 (mol)
PTHH: CH3COOH + C2H5OH --H+,to--> CH3COOC2H5 + H2O
0,12----------------------------->0,12
=> mCH3COOC2H5 = 0,12.88 = 10,56 (g)
Bài 2:
nCH3COOH = 2.0,1 = 0,2 (mol)
PTHH: 2CH3COOH + Mg --> (CH3COO)2Mg + H2
0,2------->0,1----------------------->0,1
=> mMg = 0,1.24 = 2,4 (g)
PTHH: C2H4 + H2 --to,Ni--> C2H6
0,1<--0,1
=> VC2H4(đktc) = 0,1.22,4 = 2,24 (l)
\(n_{C_2H_5OH}=\dfrac{14}{46}=\dfrac{7}{23}\left(mol\right)\)
\(C_2H_5OH+Na\rightarrow C_2H_5ONa+\dfrac{1}{2}H_2\)
\(\dfrac{7}{23}...................\dfrac{7}{23}......\dfrac{7}{46}\)
\(m_{C_2H_5ONa}=\dfrac{7}{23}\cdot68=20.7\left(g\right)\)
\(V_{H_2}=\dfrac{7}{46}\cdot22.4=3.4\left(l\right)\)
\(a) 2C_2H_5OH + 2Na \to 2C_2H_5ONa + H_2\\ n_{C_2H_5ONa} = n_{C_2H_5OH} = \dfrac{14}{46} = \dfrac{7}{23}(mol)\\ m_{C_2H_5ONa} = \dfrac{7}{23}.68 = 20,7(gam)\\ n_{H_2} = \dfrac{1}{2}n_{C_2H_5OH} = \dfrac{7}{46}(mol)\\ m_{H_2} = \dfrac{7}{46}.2 = \dfrac{7}{23}(gam)\\ b) V_{H_2} = \dfrac{7}{46}.22,4 = 3,41(lít)\)