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2CH3COOH + Zn -- > (CH3COOH)2Zn + H2
nH2 = 2,24 / 22,4 = 0,1 (mol)
=> nCH3COOH = 0,2 (mol)
mZn = 0,1. 65 = 6,5 (g)
mH2 = 0,1.2 = 0,2 (g)
mdd = 300 + 6,5 - 0,2 = 306,3 (g)
mCH3COOH = 0,2 . 60 = 12 (g)
=> C%CH3COOH = ( 12.100 ) / 306,3 = 4%
m(CH3COO)2Zn = 0,1 . 183 = 18,3 (g)
=> (18,3.100) / 306,3 = 6%
a, \(n_{CH_3COOH}=0,2.1=0,2\left(mol\right)\)
PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)
Theo PT: \(n_{Mg}=n_{H_2}=\dfrac{1}{2}n_{CH_3COOH}=0,1\left(mol\right)\)
\(\Rightarrow m=m_{Mg}=0,1.24=2,4\left(g\right)\)
\(V=V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)
Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,2\left(mol\right)\)
\(\Rightarrow m_{C_2H_5OH}=0,2.46=9,2\left(g\right)\)
\(\Rightarrow V_{ddC_2H_5OH}=\dfrac{9,2}{0,8}=11,5\left(ml\right)\)
nKOH = 0,5.0,3 = 0,15 mol
CH3COOH + KOH → CH3COOK + H2O
0,15 0,15 0,15 mol
a) CM CH3COOH = 0,15/0,2 =0,75M
b) Thể tích của dung dịch thu được sau phản ứng: 500 ml
CM CH3COOK = 0,15/0,5 = 0,3M
c) Phản ứng lên men giấm
C2H5OH + O2 → CH3COOH + H2O
0,15 0,15
→ mC2H5OH = 0,15.46 = 6,9 gam
\(n_{KOH}=0,5\cdot0,3=0,15mol\)
\(CH_3COOH+KOH\rightarrow CH_3COOK+H_2O\)
0,15 0,15 0,15 0,15
a)\(C_{M_{CH_3COOH}}=\dfrac{0,15}{0,2}=0,75M\)
b)\(C_{M_{CH_3COOK}}=\dfrac{0,15}{0,2+0,3}=0,3M\)
\(a) Zn +2 CH_3COOH \to (CH_3COO)_2Zn + H_2\\ b) n_{H_2} = n_{Zn} = \dfrac{13}{65} = 0,2(mol)\\ V_{H_2} = 0,2.22,4 = 4,48(lít)\\ n_{CH_3COOH} = 2n_{Zn} = 0,4(mol) \Rightarrow V_{dd\ CH_3COOH} = \dfrac{0,4}{1} = 0,4(lít)\\ c) C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O\\ n_{C_2H_5OH\ pư} = n_{CH_3COOH} = 0,4(mol)\\ m_{C_2H_5OH\ cần dùng} = \dfrac{0,4.46}{90\%} = 20,44(gam)\)
\(n_{BaCO_3}=\dfrac{19,7}{197}=0,1\left(mol\right)\\ a,PTHH:BaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ba+CO_2+H_2O\\ n_{CO_2}=n_{\left(CH_3COO\right)_2Ba}=n_{BaCO_3}=0,1\left(mol\right)\\ b,V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Ba}=255.0,1=25,5\left(g\right)\\ d,C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ n_{CH_3COOH}=0,1.2=0,2\left(mol\right);n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ n_{C_2H_5OH\left(TT\right)}=0,2.75\%=0,15\left(mol\right)\\ m_{C_2H_5OH\left(TT\right)}=0,15.46=6,9\left(g\right)\)
a) n glucozo = 54/180 = 0,3(mol)
n glucozo pư = 0,3.80% = 0,24(mol)
$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 +2 C_2H_5OH$
n C2H5OH = 2n glucozo = 0,48(mol)
m C2H5OH = 0,48.46 = 22,08(gam)
b)
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = n C2H5OH = 0,48(mol)
C% CH3COOH = 0,48.60/500 .100% = 5,76%
\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)
a) n Zn = 6,5/65 = 0,1(mol)
Zn + 2CH3COOH $\to$ (CH3COO)2Zn + H2
Theo PTHH :
n CH3COOH = 2n Zn =0,2(mol)
C% CH3COOH = 0,2.60/200 .100% = 6%
b) n H2 = n Zn = 0,1(mol)
=> m dd sau pư = 6,5 + 200 - 0,1.2 = 206,3 gam
Theo PTHH : n (CH3COO)2Zn = n Zn = 0,1(mol)
=> C% (CH3COO)2Zn = 0,1.183/206,3 .100% = 8,87%
c)
C2H5OH + O2 $\xrightarrow{men\ giấm}$ CH3COOH + H2O
n C2H5OH pư = n CH3COOH = 0,2(mol)
=> m C2H5OH cần dùng = 0,2.46/80% = 11,5 gam
a) nZn=0,1(mol)
PTHH: Zn + 2 CH3COOH -> (CH3COO)2Zn + H2
0,1_______0,2_________0,1_____________0,1(mol)
mCH3COOH=0,2.60=12(g)
=> C%ddCH3COOH=(12/200).100=6%
b) mdd(CH3COO)2Zn= 6,5+200-0,1.2=206,3(g)
m(CH3COO)2Zn= 183 x 0,1=18,3(g)
=>C%dd(CH3COO)2Zn= (18,3/206,3).100=8,871%
c) C2H5OH + O2 -men giấm-> CH3COOH + H2O
nC2H5OH(LT)=nCH3COOH=0,2(mol)
=> nC2H5OH(TT)=0,2 : 80%= 0,25(mol)
=>mC2H5OH=0,25 x 46= 11,5(g)