a, \(n_{K_2CO_3}=\dfrac{2,76}{138}=0,02\left(mol\right)\)

PT: \(2CH_3COOH+K_2CO_3\rightarrow2CH_3COOK+CO_2+H_2O\)

Theo PT: \(n_{CH_3COOH}=2n_{K_2CO_3}=0,04\left(mol\right)\)

\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,04.60}{50}.100\%=4,8\%\)

b, \(C_2H_5OH+O_2\underrightarrow{mengiam}CH_3COOH+H_2O\)

Theo PT: \(n_{C_2H_5OH}=n_{CH_3COOH}=0,04\left(mol\right)\Rightarrow m_{C_2H_5OH}=0,04.46=1,84\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{1,84}{0,8}=2,3\left(ml\right)\)

\(\Rightarrow V_{C_2H_5OH\left(8^o\right)}=\dfrac{2,3}{8}.100=28,75\left(ml\right)\)

\(n_{Na2CO3}=\dfrac{10,6}{106}=0,1\left(mol\right)\)

\(2CH_3COOH+Na_2CO_3\rightarrow2CH3OONa+CO_2+H_2O\)

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

a) \(V_{CO2\left(dktc\right)}=0,1.22,4=2,24\left(l\right)\)

  \(C_{MCH3COOH}=\dfrac{0,1}{0,2}=0,5\left(M\right)\)

b) \(C_{MCH3COONa}=\dfrac{0,2}{0,2}=1\left(M\right)\)

 Chúc bạn học tốt

a)

$V_{C_2H_5OH} = 200.\dfrac{11,5}{100} = 23(ml)$
$m_{C_2H_5OH} = D.V = 0,8.23 = 18,4(gam)$
$n_{C_2H_5OH} = \dfrac{18,4}{46} = 0,4(mol)$

b)

$n_{C_2H_5OH\ pư} = 0,4.80\% = 0,32(mol)$
$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
$n_{CH_3COOH} = n_{C_2H_5OH\ pư} = 0,32(mol)$
$C_{M_{CH_3COOH}} = \dfrac{0,32}{0,2} = 1,6M$

\(2CH_3COOH+CuO\rightarrow\left(CH_3COO\right)_2Cu+H_2O\\ n_{CuO}=\dfrac{37}{80}\left(mol\right)=n_{\left(CH_3COO\right)_2Cu}\\ n_{CH_3COOH}=2.\dfrac{37}{80}=\dfrac{37}{40}\left(mol\right)\\ V_{ddCH_3COOH}=\dfrac{\dfrac{37}{40}}{2}=\dfrac{37}{80}\left(l\right)\\ C_{Mdd\left(CH_3COO\right)_2Cu}=\dfrac{\dfrac{37}{80}}{\dfrac{37}{80}}=1\left(M\right)\)

Fe+2CH3COOH->(CH3COO)2Fe+H2

33\112-33\56---------------33\112

n Fe=\(\dfrac{33}{112}\) mol

=>m (CH3COO)2Fe=\(\dfrac{33}{112}\).174=51,267g

=>VCH3COOH=\(\dfrac{\dfrac{33}{56}}{3}=0,196l\)

\(n_{Fe}=\dfrac{16,5}{56}=0,29mol\)

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

0,29           0,58                      0,29            0,29       ( mol )

\(V_{H_2}=0,29.22,4=6,496l\)

\(m_{\left(CH_3COO\right)_2Fe}=0,29.174=50,46g\)

\(C_{M_{CH_3COOH}}=\dfrac{0,29}{0,3}=0,96M\)

$n_{NaOH} = \dfrac{50.10\%}{40} = 0,125(mol)$
$CH_3COOH + NaOH \to CH_3COONa + H_2O$
Theo PTHH : 

$n_{CH_3COOH} = n_{CH_3COONa} = n_{NaOH} = 0,125(mol)$
$m_{dd\ CH_3COOH} = \dfrac{0,125.60}{8\%} = 93,75(gam)$
$m_{dd\ sau\ pư} = m_{dd\ CH_3COOH} + m_{dd\ NaOH} = 143,75(gam)$
$C\%_{CH_3COONa} = \dfrac{0,125.82}{143,75}.100\% = 7,13\%$

\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ V_{C_2H_5OH}=10000.8:100=800\left(ml\right)\\ m_{C_2H_5OH}=0,8.800=640\left(g\right)\\ m_{CH_3COOH}=\dfrac{60}{46}.640.80\%=\dfrac{30720}{46}\left(g\right)\\ m_{10lethanol}=640+9200.1=9840\left(g\right)\\ m_{O_2}=\dfrac{640.32}{46}=\dfrac{20480}{46}\left(g\right)\\ C\%_{ddCH_3COOH}=\dfrac{\dfrac{30720}{46}}{9840+\dfrac{20480}{46}}.100\%\approx6,493\%\)

\(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\\ a,n_{CH_3COOH}=2.n_{H_2}=2.0,3=0,6\left(mol\right)\\ m_{ddCH_3COOH}=\dfrac{0,6.60.100}{20}=180\left(g\right)\\ b,n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=n_{H_2}=0,3\left(mol\right)\\ C\%_{dd\left(CH_3COO\right)_2Zn}=\dfrac{0,3.183}{180+0,3.65-0,3.2}.100\approx27,602\%\)

\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25mol\)

\(C_6H_{12}O_6\underrightarrow{lênmen}2C_2H_5OH+2CO_2\)

0,25                       0,5                 0,5

\(C_2H_5OH+O_2\underrightarrow{mengiấm}CH_3COOH+H_2O\)

0,5                                     0,5

\(C_{M_{CH_3COOH}}=\dfrac{0,5}{1}=0,5M\)

\(n_{C_6H_{12}O_6}=\dfrac{45}{180}=0,25\left(mol\right)\)

PTHH: 

C₆H₁₂O₆ --men rượu--> 2CO₂ + 2C₂H₅OH

0,25-------------------------------------->0,5

C₂H₅OH + O₂ --men giấm--> CH₃COOH + H₂O

0,5------------------------------------->0,5

\(\rightarrow C_{M\left(CH_3COOH\right)}=\dfrac{0,5}{1}=0,5M\)

bình \(1\) chứa số lít \(H_2SO4\) nguyên chất là : \(10.70\%=7l\)

gọi số lít từ bình \(2\) cần đổ sang là \(a\left(l\right)\)

có : sau khi đổ thì bình 1 có nồng độ \(H_2SO4\) là \(80\%\)

\(\Rightarrow\) \(\dfrac{7+0,9a}{10+a}=0,8\) \(\Leftrightarrow\) \(\left\{{}\begin{matrix}7+0,9a=0,8\left(10+a\right)\\7+0,9a=8+0,8a\\a=10\left(l\right)\end{matrix}\right.\) \(\Rightarrow\) \(a=10\) \(lít\)