Bạn xem lại xem đề cho pư với NaHCO₂ hay Na₂CO₃ nhé.

a, \(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\)

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

Theo PT: \(n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\)

\(\Rightarrow a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\)

b, \(m_{dd}=\dfrac{15}{2\%}=750\left(g\right)\)

\(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\\ NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\\a, n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\\ a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\\ b,m_{ddCH_3COOH}=\dfrac{15.100}{2}=750\left(g\right)\)

chj kb vs em nha em ib rồi ạ

\(m_{CH_3COOH}=24\%.150=36\left(g\right)\\ \rightarrow n_{CH_3COOH}=\dfrac{36}{60}=0,6\left(mol\right)\)

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

                0,6                  0,3                  0,6                  0,3 

=> V_CO2 = 0,3.22,4 = 6,72 (l)

\(m_{Na_2CO_3}=0,3.31,8\left(g\right)\)

=> \(m_{ddNa_2CO_3}=\dfrac{31,8}{21,2\%}=150\left(g\right)\)

m_CO2 = 0,3.44 = 13,2 (g)

\(m_{dd}=150+150-13,2=286,8\left(g\right)\)

\(m_{CH_3COONa}=0,3.82=24,6\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{24,6}{286,8}=8,58\%\)

Bài 1:

PTHH: \(C_2H_5OH+O_2\xrightarrow[]{mengiấm}CH_3COOH+H_2O\)

Ta có: \(n_{C_2H_5OH}=\dfrac{115\cdot0,8}{46}=2\left(mol\right)=n_{CH_3COOH\left(lýthuyết\right)}\)

\(\Rightarrow m_{CH_3COOH\left(thực\right)}=2\cdot60\cdot90\%=108\left(g\right)\)

Bài 2:

PTHH: \(C_2H_5OH+CH_3COOH\xrightarrow[H_2SO_4\left(đ\right)]{t^o}CH_3COOC_2H_5+H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{CH_3COOH}=\dfrac{60}{60}=1\left(mol\right)\\n_{C_2H_5OH}=\dfrac{92}{46}=2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\) Rượu còn dư, Axit p/ứ hết

\(\Rightarrow n_{CH_3COOC_2H_5\left(lýthuyết\right)}=1\left(mol\right)\) \(\Rightarrow m_{CH_3COOC_2H_5\left(thực\right)}=1\cdot88\cdot80\%=70,4\left(g\right)\)

a) Na₂CO₃ + 2CH₃COOH --> 2CH₃COONa + CO₂ + H₂O

b) \(n_{CH_3COOH}=\dfrac{25.6\%}{60}=0,025\left(mol\right)\)

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

              0,0125<-----0,025------------>0,025------>0,0125

=> \(m_{Na_2CO_3}=0,0125.106=1,325\left(g\right)\)

c) \(m_{dd.sau.pư}=1,325+25-0,0125.44=25,775\left(g\right)\)

\(C\%_{dd.CH_3COONa}=\dfrac{0,025.82}{25,775}.100\%=7,95\%\)

m CH3COOH=1,5g=>n=0,025 mol

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

0,025--------------0,0125----------0,025

=>m Na2CO3=0,0125.106=1,325g

=>mdd=25g

c) 

C% =\(\dfrac{0,025.82}{25+25}100=4,1\%\)

\(n_{H_2SO_4}=0.25\cdot2=0.5\left(mol\right)\)

\(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O\)

\(0.5..............0.5...............0.5\)

\(m_{Na_2CO_3}=0.5\cdot106=53\left(g\right)\)

\(C_{M_{Na_2SO_4}}=\dfrac{0.5}{0.25}=2\left(M\right)\)

\(\)250ml = 0,25l

\(n_{H2SO4}=2.0,25=0,5\left(mol\right)\)

Pt : \(Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+CO_2+H_2O|\)

            1                1                 1            1          1

          0,5               0,5              0,5

\(n_{Na2CO3}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

⇒ \(m_{Na2CO3}=0,5.106=53\left(g\right)\)      

\(n_{Na2SO4}=\dfrac{0,5.1}{1}=0,5\left(mol\right)\)

\(C_{M_{Na2SO4}}=\dfrac{0,5}{0,25}=2\left(M\right)\)

 Chúc bạn học tốt

\(CH_3COOH+NaCl\rightarrow CH_3COONa+HCl\)

\(2CH_3COOH+CaCO_3\rightarrow\left(CH_3COO\right)_2Ca+H_2O+CO_2\)

   2                         1                     1                      1         1    (mol)

 0,08                    0,04                0,04              0,04         0,04 (mol)

\(nCO_2=\dfrac{0,896}{22,4}=0,04\left(mol\right)\)

\(mCaCO_3=0,04.100=4\left(g\right)\)

=> \(mNaCl=12,5-4=8,5\left(g\right)\)

( không thấy hh B ) 

c ) .

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

 1             2               1                  1     (mol)

0,04       0,08           0,04        0,04 (mol)

\(mNa_2CO_3=0,04.106=4,24\left(g\right)\)

\(mNa_2CO_{3\left(thựctế\right)}=\)\(\dfrac{4,24.85\%}{100\%}=3,604\left(g\right)\)