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\(PTHH:2CH_3COOH+Ca\rightarrow\left(CH_3COO\right)_2Na+H_2\)
Ta có:
\(n_{\left(CH3COO\right)2Na}=\frac{4,47}{158}=0,03\left(mol\right)\)
\(\Rightarrow n_{CH3COOH}=0,03.2=0,06\left(mol\right)\)
\(\Rightarrow CM_{CH3COOH}=\frac{0,06}{0,2}=0,3M\)
\(n_{H2}=0,03\left(mol\right)\Rightarrow V_{H2}=0,03.22,4=6,72\left(l\right)\)
PTHH :
\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)
_0,06________0,06_____________________
\(\Rightarrow V_{NaOH}=\frac{0,06}{0,5}=0,12\left(l\right)=120\left(ml\right)\)
2.
CH3COOH + NaOH -> CH3COONa + H2O (1)
nNaOH=0,02(mol)
Từ 1:
nNaOH=nCH3COOH=0,02(mol)
CM axit=\(\dfrac{0,02}{0,1}=0,2M\)
a)
$C_2H_5OH + CH_3COOH \buildrel{{H_2SO_4,t^o}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O$
b)
n CH3COOC2H5 = n C2H5OH = 9,2/46 = 0,2(mol)
=> m este = 0,2.88 = 17,6 gam
c)
n este = 8,8/88 = 0,1(mol)
=> n C2H5OH = n CH3COOH = 0,1/60% = 1/6 mol
=> m C2H5OH = 46 . 1/6 = 7,67(gam) ; m CH3COOH = 60 . 1/6 = 10(gam)
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)
PTHH :
\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\uparrow\)
0,04 0,02 0,02
\(a,C_M=\dfrac{n}{V}=\dfrac{0,04}{0,1}=0,4M\)
\(b,V_{H_2}=0,02.22,4=0,448\left(l\right)\)
\(c,PTHH:\)
\(CH_3COOH+C_2H_5OH\underrightarrow{t^o,H_2SO_{4\left(đ\right)}}CH_3COOC_2H_5+H_2O\)
0,04 0,04
\(m_{este}=0,04.90\%.88=3,168\left(g\right)\)
Cảm ơn bạn nhìu nhoa ^-^