a, C\(_2\)H\(_5\)OH + O\(_2\) →(t\(^0\)) CH\(_3\)COOH + H\(_2\)O

b, V\(_r\) = \(\frac{500\cdot10}{100}\) = 50(ml)

V\(_{nc}\) = V\(_{hh}\) - V\(_r\) = 500 - 50 = 450 (ml)

⇒ m\(_{nc}\) = D\(_{nc}\) * V\(_{nc}\) = 450 * 1 = 450 (gam)

m\(_r\) = D\(_r\) * V\(_r\) = 0,78 * 50 = 39 (gam)

C\(_2\)H\(_5\)OH + O\(_2\) →(t\(^0\)) CH\(_3\)COOH + H\(_2\)O

(gam) 46 32 81

đề (gam pư) : (39*80%) → y → x

⇒ m\(_{CH_3COOH_{\left(tte\right)}}\) = x = \(\frac{39\cdot80\left(\%\right)\cdot81}{46}\) \(\sim\) 54, 94(gam)

m\(_{rượu}còn\) = 39 - (39* 80%) = 7,8 (gam)

m\(_{O_2\left(PƯ\right)}\) = y = \(\frac{39\cdot\left(80\%\right)\cdot32}{46}\) \(\sim\) 21,7 (gam)
m\(_{ddsau}\) = m\(_{dbđ}\) + m\(_{O_2\left(PƯ\right)}\) = m \(_r\) + m\(_{nc}\) + m\(_{O_2\left(PƯ\right)}\) = 39 + 450 + 21,7 = 510,7 (gam)

dung dịch sau pư có axit axetic và rượu etylic dư

C\(_{\%_rcòn}\) = \(\frac{7,8\cdot100\%}{510,7}\) \(\sim\) 1,53 (%)
C\(_{\%_{CH_3COOH}}\) = \(\frac{54,94\cdot100\%}{510,7}\) \(\sim\) 10,76(%)

\(a,m_{Na_2CO_3}=\dfrac{500.20}{100}=100\left(g\right)\\ \rightarrow n_{Na_2CO_3}=\dfrac{100}{106}=\dfrac{50}{53}\left(mol\right)\)

PTHH: \(Na_2CO_3+2CH_3COOH\rightarrow2CH_3COONa+CO_2\uparrow+H_2O\)

              \(\dfrac{50}{53}\)------->\(\dfrac{100}{53}\)--------------->\(\dfrac{100}{53}\)-------------->\(\dfrac{50}{53}\)

\(b,m_{axit}=\dfrac{100}{53}.60=\dfrac{6000}{53}\left(g\right)\\ c,m_{dd}=500+400-\dfrac{50}{53}.44=\dfrac{45500}{53}\left(g\right)\\ m_{CH_3COONa}=\dfrac{100}{53}.82=\dfrac{8200}{53}\left(g\right)\\ \rightarrow C\%_{CH_3COONa}=\dfrac{\dfrac{8200}{23}}{\dfrac{45500}{23}}.100\%=18,02\%\)

\(n_{AlCl_3}=\dfrac{26.7}{133.5}=0.2\left(mol\right)\)

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

\(.........0.6......0.2.......0.3\)

\(V_{H_2}=0.3\cdot22.4=6.72\left(l\right)\)

\(C\%HCl=\dfrac{0.6\cdot36.5}{300}\cdot100\%=7.3\%\)

a, \(n_K=\dfrac{0,975}{39}=0,025\left(mol\right)\)

A là khí H₂, B là CH₃COOK

PTHH: 2K + 2CH₃COOH → 2CH₃COOK + H₂

Mol:    0,025      0,025                                0,0125

b, \(C_{M_{ddCH_3COOH}}=\dfrac{0,025}{0,1}=0,25M\)

c, \(V_{H_2}=0,0125.22,4=0,28\left(l\right)\)

\(n_{CO_2\left(đktc\right)}=\dfrac{10,64}{22,4}=0,475\left(mol\right)\\a, K_2CO_3+2CH_3COOH\rightarrow2CH_3COOK+CO_2+H_2O\\ b,n_{K_2CO_3}=n_{CO_2}=0,475\left(mol\right)\\ \Rightarrow m_{K_2CO_3}=138.0,475=65,55\left(g\right)\\ n_{CH_3COOH}=0,475.2=0,95\left(mol\right)\\ C\%_{ddCH_3COOH}=\dfrac{0,95.60}{200}.100=28,5\%\)