a) \(n_{\left(CH_3COO\right)_2Mg}=\dfrac{7,1}{142}=0,05\left(mol\right)\)

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

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

=> V_H2 = 0,05.22,4 = 1,12 (l)

b) \(C_{M\left(dd.CH_3COOH\right)}=\dfrac{0,1}{0,025}=4M\)

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

\(n_{\left(CH_3COO\right)_2Zn}=\dfrac{14,2}{183}\approx0.077mol\)

Theo pt: nH₂ = n(CH₃COO)₂Zn = 0,077mol

=> VH₂ = 1,7248l

b) Theo pt: nCH₃COOH = 2n(CH₃COO)₂Zn = 0,154 mol

=> CM_CH3COOH = 0,154 : 0,25 = 0,616M

Câu 5 : 

\(n_{K2SO3}=\dfrac{15,8}{158}=0,1\left(mol\right)\)

Pt : \(K_2SO_3+2HCl\rightarrow2KCl+SO_2+H_2O|\)

           1             2             2          1            1

         0,1          0,2           0,2       0,2

a) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

\(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(m_{ddHCl}=\dfrac{7,3.100}{8}=91,25\left(g\right)\)

b) \(n_{KCl}=\dfrac{0,2.2}{2}=0,2\left(mol\right)\)

⇒ \(m_{KCl}=0,2.74,5=14,9\left(g\right)\)

\(m_{ddspu}=15,8+91,25-\left(0,1.64\right)=100,65\left(g\right)\)

\(C_{KCl}=\dfrac{14,9.100}{100,65}=14,80\)⁰/₀

c) \(n_{SO2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

\(V_{SO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

 Chúc bạn học tốt

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

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

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

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

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

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

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

\(m_{ddspứ}=150+150-0,15.44=293,4g\)

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

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

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)

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

n Zn= 19,5/65=0,3 (mol).

PTPƯ: Zn(0.3) +   HCl(0.6) ---->  ZnCl2(0.3) + H2(0,3)

mHCl=0,6.36.5=21.9(g)

a) C%HCl= 21.9/300.100%=7,3%

b) VH2=0,3.22,4=6,72(lít)

c) mH2=0,3.2=0,6(g)

mZnCl2=0,3.136=40,8(g)

mddZnCl2 =(19,5+300)-0,6=318,9(g)

C%=mZnCl2/mddZnCl2.100= 40,8/318,9.100=12,793%

\(n_{CaCO_3}=\dfrac{6}{100}=0,06mol\)

\(n_{CH_3COOH}=\dfrac{200}{60}=3,33mol\)

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

       3,33        >     0,06                                                          ( mol )

                              0,06            0,06                     0,06             ( mol )

\(V_{CO_2}=0,06.22,4=1,344l\)

\(m_{\left(CH_3COO\right)_2Ca}=0,06.158=9,48g\)

\(m_{ddspứ}=200+6-0,06.12=205,28g\)

\(C\%_{\left(CH_3COO\right)_2Ca}=\dfrac{9,48}{205,28}.100=4,61\%\)

\(n_{CaCO_3}=\dfrac{6}{100}=0,06\left(mol\right)\\ n_{CH_3C\text{OO}H}=\dfrac{200}{60}=3,3\left(G\right)\\ pthh:CaCO_3+2CH_3C\text{OO}H\rightarrow Ca\left(CH_3C\text{OO}\right)_2+H_2O+CO_2\) 
LTL : \(\dfrac{0,06}{1}< \dfrac{3,3}{2}\) 
=> CaCO3 hết 
theo pthh : \(n_{CO_2}=n_{CaCO_3}=0,06\left(mol\right)\) 
\(\Rightarrow V_{CO_2}=0,06.22,4=1,344\left(l\right)\) 
\(\Rightarrow C\%=\dfrac{6}{200}.100\%=3\%\dfrac{\dfrac{ }{ }C\dfrac{ }{ }\dfrac{ }{ }\dfrac{ }{ }\dfrac{ }{ }}{ }\%\)