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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 : \(n_{CH_3COOC_2H_5}=n_{C_2H_5OH}=0,15\left(mol\right)\)
=> \(m_{CH_3COOC_2H_5\left(pt\right)}=n.M=88.0,15=13,2\left(g\right)\)
=> \(m_{CH_3COOC_2H_5}=H.m=11,88\left(g\right)\)
\(a.HCl+NaOH\rightarrow NaCl+H_2O\)
PỨ trung hoà
\(b,n_{NaOH}=0,1.1=0,1mol\\ n_{NaCl}=n_{NaOH}=n_{HCl}0,1mol\\ m=m_{HCl}=0,1.36,5=3,65g\\ c,m_{NaCl}=0,1.58,5=5,85g\\ d,n_{HCl}=\dfrac{73.10}{100.36,5}=0,2mol\\ \Rightarrow\dfrac{0,1}{1}< \dfrac{0,2}{1}\Rightarrow HCl.dư\\ n_{HCl,pứ}=n_{NaOH}=0,1mol\\ m_{HCl,dư}=\left(0,2-0,1\right).36,5=3,65g\)
\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)
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