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\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)
nNaOH=0,1.4=0,4 mol
NaOH+CH3COOH -->CH3COONa+H2O
0,4 --> 0,4
mCH3COOH=0,4.60=24g
b/
CH3COOH+C2H5OH --->CH3COOCH2CH3+H2O
0,4. --> x mol
Chỉ phản ứng 60% ta có : x=0,4.60/100=0,24mol
meste=0,24.88=21,12g
\(n_{CH_3COOH}=\dfrac{300\cdot5\%}{60}=0.25\left(mol\right)\)
\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
\(0.25........................................................0.125\)
\(V_{H_2}=0.125\cdot22.4=2.8\left(l\right)\)
\(n_{C_2H_5OH}=0.1\cdot2=0.2\left(mol\right)\)
\(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\left(ĐK:H_2SO_{4\left(đ\right)},t^0\right)\)
\(0.2......................0.2.....................0.2\)
\(\Rightarrow CH_3COOHdư\)
\(m_{CH_3COOC_2H_5}=0.2\cdot88=17.6\left(g\right)\)
a) n CH3COOH = 300.5%/60 = 0,25(mol)
Zn + 2CH3COOH $\to$ (CH3COO)2Zn + H2
Theo PTHH :
n H2 = 1/2 n CH3COOH = 0,25/2 = 0,125(mol)
V H2 = 0,125.22,4 = 2,8(lít)
b) n C2H5OH = 0,1.2 = 0,2(mol)
\(CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\)
Ta thấy :
n CH3COOH = 0,25 > n C2H5OH = 0,2 => CH3COOH dư
n CH3COOC2H5 = n C2H5OH = 0,2 mol
=> m CH3COOC2H5 = 0,2.88 = 17,6 gam
a) \(n_{CH_3COOH}=\dfrac{30}{60}=0,5\left(mol\right)\)
PTHH: CH3COOH + C2H5OH --H2SO4(đ),to--> CH3COOC2H5 + H2O
0,5------->0,5------------------------>0,5
=> mC2H5OH = 0,5.46 = 23 (g)
b) mCH3COOC2H5 = 0,5.88 = 44 (g)
PTHH : \(C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\)
\(n_{C_2H_5OH}=\frac{m}{M}=\frac{6,9}{46}=0,15\left(mol\right)\)
\(m_{ddCH_3COOH}=D.V=100.1,05=105\left(g\right)\)
Lại có : \(C\%_{CH_3COOH}=\frac{m_{CH_3COOH}}{m_{dd}}.100\%=36\%\)
=> \(m_{CH_3COOH}=37,8\left(g\right)\)
=> \(n_{CH_3COOH}=\frac{m}{M}=\frac{37,8}{60}=0,63\left(mol\right)\)
Ta có : \(\frac{n_{C_2H_5OH}}{n_{CH_3COOH}}=\frac{0,15}{0,63}~0,23< 1\)
=> C2H5OH phản ứng hết, CH3COOH còn dư .
- Theo PTHH : \(n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,15\left(mol\right)\)
=> \(m_{CH_3COOC_2H_5\left(pt\right)}=n.M=88.0,15=13,2\left(g\right)\)
=> \(m_{CH_3COOC_2H_5}=H.m=11,88\left(g\right)\)