áp dụng sơ đồ đường chéo ta có:

\(\Rightarrow\) \(\frac{m1}{m2}\)= \(\frac{40}{10}\)= \(\frac{4}{1}\)

\(\Rightarrow\) \(\frac{200}{m2}\)= \(\frac{4}{1}\)

\(\Rightarrow\) m₂= \(\frac{200}{4}\)= 50( g)

vậy cần thêm 50g nước vào 200g dd CH₃COOH 50% để thu được dd CH₃COOH 40%

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

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

\(n_{MgCO_3}=\dfrac{12.6}{84}=0.15\left(mol\right)\)

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

\(0.15..........0.3............................................0.15\)

\(V_{CO_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{CH_3COOH}=0.3\cdot60=18\left(g\right)\)

\(C\%_{CH_3COOH}=\dfrac{18}{200}\cdot100\%=9\%\)

Pthh MgCO3+2CH3COOH--->(CH3COO)2Mg+CO2+H2O

Ta có nMgCO3=0,15 mol

Theo pthh thì nCH3COOH=0,3 mol

=>mCH3COOH=17,7 g

=>C%CH3COOH=8,85%

Câu a mik chưa bt tính V của cái nào

CH3COOH+NaHCO3->CH3COONa+H2O+CO2

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

n CO2=0,1 mol

=>C% axit=\(\dfrac{0,1.60}{200}.100\)=3%

m NaHCO3=0,1.84=8,4g

c) C% CH3COONa=\(\dfrac{0,1.82}{200+8,4}.100=3,934\%\)

\(n_{Fe}=\dfrac{6,5}{56}=\dfrac{13}{112}mol\)

\(m_{CH_3COOH}=\dfrac{90\cdot20\%}{100\%}=18g\Rightarrow n_{CH_3COOH}=0,3mol\)

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

\(\dfrac{13}{112}\)       0,3                       0                        0

\(\dfrac{13}{112}\)       \(\dfrac{13}{56}\)                       \(\dfrac{13}{112}\)                   \(\dfrac{13}{112}\)

   0       \(\dfrac{19}{280}\)                      \(\dfrac{13}{112}\)                  \(\dfrac{13}{112}\)

a)\(m_{\left(CH_3COO\right)_2Fe}=\dfrac{13}{112}\cdot174=20,2g\)

\(m_{H_2}=\dfrac{13}{112}\cdot2=\dfrac{13}{56}g\)

\(m_{dd\left(CH_3COO\right)_2Fe}=6,5+90-\dfrac{13}{56}=96,27g\)

\(C\%=\dfrac{20,2}{96,27}\cdot100\%=20,98\%\)

a, \(m_{CH_3COOH}=20.3,75\%=0,75\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{0,75}{60}=0,0125\left(mol\right)\)

PT: \(2CH_3COOH+Ca\left(OH\right)_2\rightarrow\left(CH_3COO\right)_2Ca+2H_2O\)

Theo PT: \(n_{Ca\left(OH\right)_2}=\dfrac{1}{2}n_{CH_3COOH}=0,00625\left(mol\right)\)

\(\Rightarrow V_{ddCa\left(OH\right)_2}=\dfrac{0,00625}{0,2}=0,03125\left(l\right)=31,25\left(ml\right)\)

b, \(n_{\left(CH_3COO\right)_2Ca}=\dfrac{1}{2}n_{CH_3COOH}=0,00625\left(mol\right)\)

\(\Rightarrow m_{\left(CH_3COO\right)_2Ca}=0,00625.158=0,9875\left(g\right)\)

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

Ta có: \(n_{CH_3COONa}=\dfrac{9,84}{82}=0,12\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=n_{NaOH}=n_{CH_3COONa}=0,12\left(mol\right)\)

\(\Rightarrow V_{ddCH_3COOH}=\dfrac{0,12}{0,5}=0,24\left(l\right)\)

\(m_{NaOH}=0,12.40=4,8\left(g\right)\Rightarrow m_{ddNaOH}=\dfrac{4,8}{20\%}=24\left(g\right)\)