a) nCH3COOH= 0,4(mol)

PTHH: CH3COOH + NaOH -> CH3COONa + H2O

0,4____________0,4(mol)

=> mNaOH=0,4. 40=16(g)

b) nCH3COOH= 1(mol)

nC2H5OH= 100/46= 50/23(mol)

Vì : 1/1< 50/23 :1

=> C2H5OH dư, CH3COOH hết, tính theo nCH3COOH.

PTHH: CH3COOH + C2H5OH \(⇌\) CH3COOC2H5 + H2O (đk: H⁺ , nhiệt độ)

Ta có: nCH3COOC2H5(thực tế)= 0,625(mol)

Mà theo LT: nCH3COOC2H5(LT)= nCH3COOH=1(mol)

=>H= (0,625/1).100=62,5%

\(a)n_{CH_3COOH} = 0,2.2 = 0,4(mol)\\ Mg + 2CH_3COOH \to (CH_3COO)_2Mg + H_2\\ n_{Mg} = \dfrac{1}{2}n_{CH_3COOH} = 0,2(mol)\\ m_{Mg} = 0,2.24 = 4,8(gam)\\ b)\\ CH_3COOH + C_2H_5OH \buildrel{{H_2SO_4}}\over\rightleftharpoons CH_3COOC_2H_5 + H_2O\\ n_{CH_3COOH\ pư} = n_{este} = \dfrac{24,64}{88} = 0,28(mol)\\ H = \dfrac{0,28}{0,4}.100\% = 70\%\)

a, Ta có: \(n_{CH_3COOH}=\dfrac{9,6}{60}=0,16\left(mol\right)\)

PT: \(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

Theo PT: \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{1}{2}n_{CH_3COOH}=0,08\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Mg}=0,08.142=11,36\left(g\right)\)

b, PT: \(CH_3COOH+C_2H_5OH\underrightarrow{t^o,xt}CH_3COOC_2H_5+H_2O\)

Theo PT: \(n_{CH_3COOC_2H_5\left(LT\right)}=n_{CH_3COOH}=0,16\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5\left(LT\right)}=0,16.88=14,08\left(g\right)\)

Mà: thực tế thu được 10,56 (g)

\(\Rightarrow H\%=\dfrac{10,56}{14,08}.100\%=75\%\)

2C2H5OH+Na->2C2H5ONa +H2

0,3------------------------------------0,15

2CH3COOH+Na->2CH3COONa+H2

0,1-------------------------------------->0,05

NaOH+CH3COOH->CH3COONa+H2O

0,1-------0,1 mol

n khí =4,48 \22,4=0,2 mol

n NaOH=0,5.0,2=0,1 mol

=>nH2 pt2=0,05 

=>n H2 pt1=0,15

=>mC2H5OH=0,3.46=13,8g

=>m CH3COOH=0,1.60=6g

a.b.\(n_{Mg}=\dfrac{4,8}{24}=0,2mol\)

\(2Mg+2CH_3COOH\rightarrow2\left(CH_3COO\right)_2Mg+H_2\)

0,2             0,2                          0,2                            ( mol )

\(C_{M_{CH_3COOH}}=\dfrac{0,2}{0,2}=1M\)

\(m_{\left(CH_3COO\right)_2Mg}=0,2.142=28,4g\)

c.Sửa đề: thu được 9,2g este

\(n_{CH_3COOC_2H_5}=\dfrac{9,2}{88}=0,1mol\)

             \(CH_3COOH+C_2H_5OH\rightarrow CH_3COOC_2H_5+H_2O\)

Thực tế:       0,2                                           0,1                     ( mol )

Lý thuyết:      0,1                                              0,1                 ( mol )

\(H=\dfrac{0,1}{0,2}.100=50\%\)

\(a,m_{CH_3COOH}=\dfrac{200.10}{100}=20\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{20}{60}=\dfrac{1}{3}\left(mol\right)\)

PTHH: 2CH₃COOH + Na₂CO₃ ---> 2CH₃COONa + CO₂ + H₂O

               \(\dfrac{1}{3}\)-------------->\(\dfrac{1}{6}\)

b, => m_Na2CO3 = \(\dfrac{1}{6}.106=\dfrac{53}{3}\left(g\right)\)

tk

1313-------------->16.106=533(g)

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)

nZn=6,5/65=0,1

PTHH:Zn+ 2CH3COOH=(CH3COO)2Zn + H2O

           0,1    0,2                   0,1

mCH3COOH=0,2.60=12(g)

mddCH3COOH=12+200=212(g)

C%ddCH3COOH=mct/mdd .100%=12/212 .100%=5,7%

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