a)

$CH_3COOH + NaHCO_3 \to CH_3COONa + CO_2 + H_2O$

b)

n NaHCO3 = n CH3COOH = 100.12%/60 = 0,2(mol)

m dd NaHCO3 = 0,2.84/8% = 210(gam)

c)

n CO2 = n CH3COOH = 0,2(mol)

=> V CO2 = 0,2.22,4 = 4,48(lít)

d)

m dd = m dd CH3COOH + m dd NaHCO3 - m CO2 = 100  + 210 - 0,2.44 = 301,2(gam)

C% CH3COONa = 0,2.82/301,2   .100% = 5,44%

Ta có: \(n_{NaOH}=0,1.0,5=0,05\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COOH}=n_{CH_3COONa}=n_{NaOH}=0,05\left(mol\right)\)

a, \(C_{M_{CH_3COOH}}=\dfrac{0,05}{0,2}=0,25\left(M\right)\)

b, \(m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)

a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PT: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

b, \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\)

c, \(n_{CH_3COOH}=2n_{Zn}=0,2\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{60\%}=20\left(g\right)\)

\(m_{CH_3COOH}=\dfrac{80.9}{100}=7,2\left(g\right)\)

\(n_{CH_3COOH}=\dfrac{7,2}{60}=0,12\left(mol\right)\)

PTHH :

\(15CH_3COOH+10NaHCO_3\rightarrow10CH_3COONa+2H_2O+20CO_2\uparrow\)

0,12                        0,08                 0,08                     0,016     0,16 

\(a,m_{NaHCO_3}=84.0,08=6,72\left(g\right)\)

\(m_{ddNaHCO_3}=\dfrac{6,72.100}{4,2}=160\left(g\right)\)

\(b,m_{CH_3COONa}=0,08.82=6,56\left(g\right)\)

\(m_{H_2O}=0,016.18=0,288\left(g\right)\)

\(m_{CO_2}=0,16.44=7,04\left(g\right)\)

\(m_{ddCH_3COONa}=80+160-0,288-7,04=232,672\left(g\right)\)

\(C\%=\dfrac{6,56}{232,672}\approx2,82\%\)

a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{10,4}.100\%\approx80,77\%\\\%m_{MgO}\approx19,23\%\end{matrix}\right.\)

b, \(n_{MgO}=\dfrac{10,4-0,1.84}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{MgO}+2n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

a) 

\(n_{CO_2}=\dfrac{3,36}{22,4}=0,15\left(mol\right)\)

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

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

=> \(C\%_{dd.CH_3COOH}=\dfrac{0,3.60}{200}.100\%=9\%\)

b)

\(m_{dd.Na_2CO_3}=\dfrac{0,15.106.100}{15}=106\left(g\right)\)

c)

PTHH: 2CH₃COOH + Ba(OH)₂ --> (CH₃COO)₂Ba + 2H₂O

                    0,3--------->0,15

=> \(V_{dd.Ba\left(OH\right)_2}=\dfrac{0,15}{0,5}=0,3\left(l\right)=300\left(ml\right)\)

CH3COOH+Na->CH3COONa+0,5H2

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

=>n CH3COOH=0,3 mol

=>m CH3COONa=0,3.82=24,6g

Bài 1:

n_CH3COOH = 0,08.1,5 = 0,12 (mol)

PTHH: CH₃COOH + C₂H₅OH --H⁺,t^o--> CH₃COOC₂H₅ + H₂O

                0,12----------------------------->0,12

=> m_CH3COOC2H5 = 0,12.88 = 10,56 (g)

Bài 2:

n_CH3COOH = 2.0,1 = 0,2 (mol)

PTHH: 2CH₃COOH + Mg --> (CH₃COO)₂Mg + H₂

                  0,2------->0,1----------------------->0,1

=> m_Mg = 0,1.24 = 2,4 (g)

PTHH: C₂H₄ + H₂ --t^o,Ni--> C₂H₆

              0,1<--0,1

=> V_C2H4(đktc) = 0,1.22,4 = 2,24 (l)

a) 

\(n_{Na_2CO_3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

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

                0,1--------->0,2------------->0,2------------>0,1

=> m_CH3COOH = 0,2.60 = 12 (g)

\(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,2}{0,4}=0,5M\)

b) \(n_{C_2H_5OH}=\dfrac{13,8}{46}=0,3\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

Xét tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,3}{1}\) => Hiệu suất tính theo CH₃COOH

\(n_{CH_3COOH\left(pư\right)}=\dfrac{0,2.80}{100}=0,16\left(mol\right)\)

PTHH: CH₃COOH + C₂H₅OH --H₂SO₄,t^o--> CH₃COOC₂H₅ + H₂O

                 0,16------------------------------------->0,16

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