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) n glucozo = 54/180 = 0,3(mol)

n glucozo pư = 0,3.80% = 0,24(mol)

$C_6H_{12}O_6 \xrightarrow{t^o} 2CO_2 +2 C_2H_5OH$

n C2H5OH = 2n glucozo = 0,48(mol)

m C2H5OH = 0,48.46 = 22,08(gam)

b)

$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$
n CH3COOH = n C2H5OH = 0,48(mol)

C% CH3COOH = 0,48.60/500  .100% = 5,76%

@buithianhtho giải giúp e với

Ta có:

\(V_{C2H5OH}=4,6.14\%=0,644\left(l\right)=644\left(ml\right)\)

\(\Rightarrow m_{C2H5OH}=0,8.644=515,2\left(g\right)\)

\(\Rightarrow n_{C2H5OH}=\frac{515,2}{46}=11,2\left(mol\right)\)

\(V_{H2O}=4,6-0,644=3,956\left(l\right)=3956\left(ml\right)\)

\(\Rightarrow m_{H2O}=3956\left(g\right)\)

Mà H = 30%

\(\Rightarrow n_{C2H5OH\left(pư\right)}=11,2.30\%=3,36\left(mol\right)\)

\(C_2H_5OH+O_2\underrightarrow{^{men.giam}}CH_3COOH+H_2O\)

\(\Rightarrow n_{O2}=n_{CH3COOH}=3,36\left(mol\right)\)

\(m_{dd\left(spu\right)}=515,2+3956+3,36.32=4578,72\left(g\right)\)

\(\Rightarrow C\%_{CH3COOH}=\frac{3,36.60}{4578,72}.100\%=4,4\%\)

a)

$C_2H_5OH + O_2 \xrightarrow{men\ giấm} CH_3COOH + H_2O$

V rượu =  57,5.12/100 = 6,9(lít) = 6900(cm3)

=> m rượu = 6900.0,8 = 5520(gam)

Theo PTHH :

n CH3COOH = n C2H5OH = 5520/46 = 120(mol)

m CH3COOH = 120.60 = 7200(gam)

b)

m dd giấm = 7200/4% = 180 000(gam)

\(V_r=57.5\cdot0.12=6.9\left(l\right)\)

\(m_{C_2H_5OH}=6.9\cdot0.8=5.52\left(g\right)\)

\(n_{C_2H_5OH}=\dfrac{5.52}{46}=0.12\left(mol\right)\)

\(n_{C_2H_5OH\left(pư\right)}=0.12\cdot92\%=0.1104\left(mol\right)\)

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

\(0.1104........................0.1104\)

\(m_{dd_{CH_3COOH}}=\dfrac{0.1104\cdot60}{4\%}=165.6\left(g\right)\)

a. \(m_{C_2H_5OH}=\dfrac{10.0,8.8}{100}=0,64\left(kg\right)\)

\(n_{C_2H_5OH}=\dfrac{0,64}{46}=\dfrac{8}{575}\left(k-mol\right)\)

\(C_2H_5OH+O_2\rightarrow\left(t^o,men.giấm\right)CH_3COOH+H_2O\)

   \(\dfrac{8}{575}\)                                                     \(\dfrac{8}{575}\)          ( k-mol )

\(m_{CH_3COOH}=\dfrac{8}{575}.60.92\%=0,768\left(kg\right)=768\left(g\right)\)

b.\(m_{dd_{CH_3COOH}}=\dfrac{768.100}{4}=19200\left(g\right)\)

a.\(V_{C_2H_5OH}=\dfrac{10.8}{100}=0,8ml\)

\(m_{C_2H_5OH}=0,8.0,8=1,6g\)

\(n_{C_2H_5OH}=\dfrac{1,6}{46}=0,034mol\)

\(C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\)

 0,034                                            0,034                ( mol )

\(m_{CH_3COOH}=0,034.60.80\%=1,632g\)

b.\(m_{dd_{CH_3COOH}}=\dfrac{1,632}{5\%}=32,64g\)