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![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi số mol mỗi phần của CH3OH, CH3COOH là x, y (mol)
P1: CH3OH + Na --> CH3ONa +1/2H2(1)
CH3COOH + Na --> CH3COONa + 1/2 H2 (2)
Từ PT (1) và (2) ta thấy:
nH2= 0.5 (x +y) = 19.04/22.4=0.85
<=> x + y= 1.7 (3)
P2:2CH3COOH + CaCO3 -->(CH3COO)2Ca + CO2 + H2O
Từ PTHH: nCH3COOH= 2 nCO2= 2 * 5.6/22.4= 0.5
=> y= 0.5 mol
Từ (3) => x= 1.3 - 0.5= 0.8 (mol)
mhh=a=mCH3OH+mCH3COOH
= 2*0.8*32 + 2*0.5*60= 111.2g
b)2CH3COOH + Ba(OH)2 --> (CH3COO)2Ba + H2O
=> nBa(OH)2= 1 (mol)
VddBa(OH)2= 1/0.25=4 (l)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
1,
- Xét phần 2:
\(n_{CH_3COOC_2H_5}=\dfrac{4,4}{88}=0,05\left(mol\right)\)
PTHH: CH3COOH + C2H5OH \(\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}\) CH3COOC2H5 + H2O
LTL: 0,5a < 0,5b (do a < b) => C2H5OH dư
Theo pthh: \(n_{CH_3COOH\left(pư\right)}=n_{CH_3COOC_2H_5}=0,05\left(mol\right)\)
Mà H = 50%
=> \(n_{CH_3COOH\left(bđ\right)}=\dfrac{0,05}{50\%}=0,1\left(mol\right)\)
Xét phần 2:
\(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PTHH:
2CH3COOH + 2Na ---> 2CH3COONa + H2
0,1--------------------------------------------->0,05
2C2H5OH + 2Na ---> 2C2H5ONa + H2
0,4<----------------------------------------0,2
=> Trong X có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=0,1.2=0,2\left(mol\right)\\n_{C_2H_5OH}=0,4.2=0,8\left(mol\right)\end{matrix}\right.\)
2, Gọi \(\left\{{}\begin{matrix}n_{CH_3COOH}=a\left(mol\right)\\n_{C_2H_5OH}=b\left(mol\right)\end{matrix}\right.\left(a,b>0\right)\)
\(n_{H_2O}=\dfrac{23,4}{18}=1,3\left(mol\right)\)
PTHH:
CH3COOH + 2O2 --to--> 2CO2 + 2H2O
a------------------------------------------>2a
C2H5OH + 3O2 --to--> 2CO2 + 3H2O
b-------------------------------------->3b
=> Hệ pt \(\left\{{}\begin{matrix}60a+46b=25,8\\2a+3b=1,3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,3\left(mol\right)\end{matrix}\right.\left(TM\right)\)
PTHH: \(CH_3COOH+C_2H_5OH\xrightarrow[t^o]{H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
LTL: 0,2 < 0,3 => Rượu dư
Theo pthh: \(n_{CH_3COOH\left(pư\right)}=n_{CH_3COOC_2H_5}=\dfrac{14,08}{88}=0,16\left(mol\right)\)
=> \(H=\dfrac{0,16}{0,2}.100\%=80\%\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
Fe + 2HCl \(\rightarrow\) FeCl2 + H2
\(n_{HCl}=\dfrac{6,72}{22,4}=0,3mol\)
\(n_{Fe}=\dfrac{0,3}{2}=0,15mol\\ m_{Fe}=0,15.56=8,4g\)
Câu 2:
Zn + H2SO4 \(\rightarrow\) ZnSO4 + H2
\(n_{H_2}=\dfrac{13,44}{22,4}=0,6mol\\ n_{ZnSO_4}=0,6mol\)
\(m_{ZnSO_4}=0,6.161=96,6g\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(n_{H_2}=n_O=0,25\left(mol\right)\\\dfrac{a}{2}=11,15-0,25.16\\ =>a=14,3\left(g\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
nC2H5OH=0,3(mol)
PTHH: C2H5OH + Na -> C2H5ONa + 1/2 H2
nH2=nC2H5OH/2=0,3/2=0,15(mol)
=>V(H2,đktc)=0,15.22,4=3,36(l)
![](https://rs.olm.vn/images/avt/0.png?1311)
- 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) \(m_{C_2H_5ONa}=n.M=0,2\times68=13,6\left(g\right)\)
\(m_{CH_3COONa}=n.M=0,4\times82=32,8\left(g\right)\)
Gọi số mol mỗi phần của CH3OH, CH3COOH là x, y (mol)
P1: CH3OH + Na --> CH3ONa +1/2H2(1)
CH3COOH + Na --> CH3COONa + 1/2 H2 (2)
Từ PT (1) và (2) ta thấy:
nH2= 0.5 (x +y) = 19.04/22.4=0.85
<=> x + y= 1.7 (3)
P2:2CH3COOH + CaCO3 -->(CH3COO)2Ca + CO2 + H2O
Từ PTHH: nCH3COOH= 2 nCO2= 2 * 5.6/22.4= 0.5
=> y= 0.5 mol
Từ (3) => x= 1.3 - 0.5= 0.8 (mol)
mhh=a=mCH3OH+mCH3COOH
= 2*0.8*32 + 2*0.5*60= 111.2g