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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)
PTHH :
\(2CH3COOH+Mg->\left(CH3COO\right)2Mg+H2\)
0,02mol................0,01mol......................................0,01mol
a) VH2(đktc) = 0,01.22,4 = 0,224(l)
mMg = 0,01.24 = 0,24(g)
b) \(CH3COOH+C2H5OH\underrightarrow{H2SO4,đặc,t0}CH3COOC2H5+H2O\)
Ta có : nC2H5OH = 1,15/46 = 0,025(mol)
Ta có : nC2H5OH = 0,025 > nCH3COOH
=> nC2H5OH còn dư
=> n(etyl axetat) = nCH3COOH = 0,02(mol)
=> m(etyl axetat) = 0,02.88 = 1,76(g)
nC2H5OH = 8.05/46 = 0.175 (mol)
nCH3COOH = 36/60 = 0.6 (mol)
nCH3COOC2H5 = 12.32/88 = 0.14 (mol)
C2H5OH + CH3COOH <-H2SO4đ,t0-> CH3COOC2H5 + H2O
1.......................1
0.175................0.6
LTL : 0.175/1 < 0.6/1
=> CH3COOH dư
mCH3COOH (dư) = ( 0.6 - 0.175) * 60 = 25.5 (g)
nCH3COOC2H5 = nC2H5OH = 0.175 (mol)
H% = 0.14/0.175 * 100% = 80%
mHCl= 3,65%.400= 14,6(g) => nHCl=14,6/36,5=0,4(mol)
a) PTHH: Mg +2 HCl -> MgCl2 + H2
0,2__________0,4_____0,2____0,2(mol)
V(H2,đktc)=0,2.22,4=4,48(l)
b)mMg=0,2.24=4,8(g)
c) mMgCl2= 0,2.95=19(g)
mddMgCl2= 400+4,8 - 0,2.2= 404,4(g)
=> C%ddMgCl2= (19/404,4).100=4,698%
\(n_{\left(CH_3COO\right)_2Mg}=\dfrac{2,84}{142}=0,02\left(mol\right)\)
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
\(m_{este}=0,04.90\%.88=3,168\left(g\right)\)
\(n_{CH_3COOH}=0,2\cdot0,1=0,02mol\)
a)\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
0,02 0,01 0,01 0,01
b)\(V_{H_2}=0,01\cdot22,4=0,224l=224ml\)
\(m_{Mg}=0,01\cdot24=0,24g\)
c)\(CH_3COOH+C_2H_5OH\xrightarrow[xtH_2SO_4đ]{t^o}CH_3COOC_2H_5+H_2O\)
0,02 \(\dfrac{1,15}{46}=0,025\) 0,02
\(m_{etylaxetat}=0,02\cdot88=1,76g\)
\(H=80\%\Rightarrow m_{CH_3COOC_2H_5}=1,76\cdot80\%=1,408g\)