Ta có: \(n_{Na_2CO_3}=0,4.1=0,4\left(mol\right)\)

PT: \(Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\)

a, \(n_{HCl}=2n_{Na_2CO_3}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,6}=\dfrac{4}{3}\left(M\right)\)

b, \(n_{NaCl}=2n_{Na_2CO_3}=0,8\left(mol\right)\Rightarrow m_{NaCl}=0,8.58,5=46,8\left(g\right)\)

\(n_{CO_2}=n_{Na_2CO_3}=0,4\left(mol\right)\Rightarrow V_{CO_2}=0,4.24,79=9,916\left(l\right)\)

c, \(C_{M_{NaCl}}=\dfrac{0,8}{0,4+0,6}=0,8\left(M\right)\)

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

=> \(V_{dd.CH_3COOH}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

b) \(m_{CH_3COONa}=0,2.82=16,4\left(g\right)\)

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

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)

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

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

\(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Fe+H_2\)

a, Theo PT: \(n_{CH_3COOH}=2n_{Fe}=0,2\left(mol\right)\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)

\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{10\%}=120\left(g\right)\)

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

b, Theo PT: \(n_{\left(CH_3COO\right)_2Fe}=n_{Fe}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 5,6 + 120 - 0,1.2 = 125,4 (g)

\(\Rightarrow C\%_{\left(CH_3COO\right)_2Fe}=\dfrac{0,1.174}{125,4}.100\%\approx13,88\%\)

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

b) \(n_{CH_3COOH}=\dfrac{200.12\%}{60}=0,4\left(mol\right)\)

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

                     0,4------->0,2------------------------->0,2

=> \(m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

=> \(m_{dd.Na_2CO_3}=\dfrac{21,2.100}{50}=42,4\left(g\right)\)

c) \(V_{CO_2}=0,2.22,4=4,48\left(l\right)\)

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

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

                  0,2--------->0,4--------------->0,4------->0,2

=> V_CO2 = 0,2.22,4 = 4,48 (l)

b) \(m_{dd.CH_3COOH}=\dfrac{0,4.60}{12\%}=200\left(g\right)\)

c) m_{dd sau pư} = 21,2 + 200 - 0,2.44 = 212,4 (g)

=> \(C\%_{muối}=\dfrac{0,4.82}{212,4}.100\%=15,44\%\)

\(n_{Al}=\dfrac{10,8}{27}=0,4\left(mol\right)\)

PTHH :

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\uparrow\)

0,4      1,2           0,4         0,6 

\(a,V_{H_2}=0,6.22,4=13,44\left(l\right)\)

\(b,m_{HCl}=1,2.36,5=43,8\left(g\right)\)

\(c,m_{AlCl_3}=133,5.0,4=53,4\left(g\right)\)

\(m_{ddHCl}=\dfrac{43,8.100}{10}=438\left(g\right)\)

\(m_{ddAlCl_3}=10,8+438-\left(0,6.2\right)=447,6\left(g\right)\)

\(C\%=\dfrac{53,8}{447,6}.100\%\approx12,02\%\)

\(m_{CH_3COOH}=150.12\%=18g\)

\(n_{CH_3COOH}=\dfrac{18}{60}=0,3mol\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

      0,3                   0,15                0,3                 0,15               ( mol )

\(m_{ddNa_2CO_3}=\left(0,15.106\right):10,6\%=150g\)

\(V_{CO_2}=0,15.22,4=3,36l\)

\(m_{CH_3COONa}=0,3.82=24,6g\)

\(m_{ddspứ}=150+150-0,15.44=293,4g\)

\(C\%_{CH_3COONa}=\dfrac{24,6}{293,4}.100=8,28\%\)